Answer:
-471 Kj/mole acrylic acid
Explanation:
THIS IS THE COMPLETE QUESTION BELOW;
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?
The two equations from the reaction can be written as;
a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)
Δ H= -414Kj ........................ equation (a)
b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)
In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water
Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.
6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)
Δ H= -2484Kj.................. equation (c)
6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj
Then add equation (c) and equation(b) then we have
6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj
ΔH(net)= -2352Kj/5moles
=-471Kj/mole
therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid
Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?
N2O4(g) ⇌ 2NO2(g)
a. [NO]^2 [N2O4]
b. [NO2]/ [N2O4]^2
c. [NO2]/[N2O4]
d. [NO2]^2/[N2O4]
Answer:
D. [NO₂]²/[N₂O₄]
Explanation:
The equilibrium constant expression for a reaction is products over reactants. Since NO₂ has a coefficient of 2, it will become an exponent.
So, it would be:
[NO₂]²/[N₂O₄]
Hope that helps.
The correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide (N₂O₄) and nitrogen dioxide (NO₂) is option d. [NO₂]²/[N₂O₄].
In the balanced equation N₂O₄(g) ⇌ 2NO₂(g), the stoichiometric coefficients indicate that two moles of NO₂ are produced for every one mole of N₂O₄ consumed.
The equilibrium constant expression is derived from the molar concentrations of the species involved in the equilibrium. In this case, the expression is written as [NO₂]²/[N₂O₄], where [NO₂] represents the molar concentration of NO₂ and [N₂O₄] represents the molar concentration of N₂O₄.
By squaring the concentration of NO₂ and dividing it by the concentration of N₂O₄, the equilibrium-constant expression correctly accounts for the stoichiometry of the reaction and the relative concentrations of the species involved.
Hence, the correct option is option d.
Learn more about equilibrium here:
https://brainly.com/question/30694482
#SPJ 4
Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.
Answer:
Q = 246 kJ
Explanation:
It is given that,
Mass of water, m = 200 g
Let initial temperature, [tex]T_i=5^{\circ}[/tex]
Final temperature of water, [tex]T_f=300^{\circ} C[/tex]
We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]
So, the heat energy needed to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]
or
Q = 246 kJ
So, the heat energy of 246 kJ is needed.
At 298 K, Kc is 2.2×105 for the reaction F(g)+O2(g)⇌O2F(g) . What is the value of Kp at this temperature? Express your answer using two significant figures.
Answer:
The value of Kp at this temperature is 9.0*10³
Explanation:
The equilibrium constant Kp describes the relationship that exists between the partial pressures of the reactants and products, while Kc represents the relationship that exists between the concentrations of the reactants and products that participate in the reaction.
The general relationship between the constants Kp and Kc results:
Kp=Kc*[tex](R*T)^{moles of product - moles of reagent}[/tex]
In this case:
Kc= 2.2*10⁵R = gas constant = 0.0821 [tex]\frac{atm*L}{mol*K}[/tex] T = Kelvin temperature = 298 K moles of gaseous products - moles of gaseous reactants = 1 - 2 = -1Replacing:
Kp=2.2*10⁵*[tex](0.0821*298)^{-1}[/tex]
Solving:
Kp≅9.0*10³
The value of Kp at this temperature is 9.0*10³
Which of the choices is not a statement or direct application of the second law of thermodynamics? There are no 100% efficient heat engines. The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system. Heat energy naturally transfers from a high temperature substance to a low temperature substance. All natural processes tend toward states of increased entropy.
Answer:
The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics.
Explanation:
Second law of thermodynamics states that heat can be transfer spontaneously from high temperature to low temperature only.
The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics because according to the second law of thermodynamics it is impossible in any system for heat transfer from a source to completely convert to work done in a cyclical process( bring the system to its original stage after each cycle) in which the system then return to it's original stage.
Which molecule is an aromatic hydrocarbon? Two central carbons are double bonded to each other; the pair is single bonded to C H 3 above left and right, and to H below left and right. A skeletal model has line segments that slant up, down, up, and up again in a triple bond. Two hexagon rings with carbons at each corner have alternating double bonds and share one side. H is single bonded to all the C's except the ones on the shared side. Two carbons are triple bonded to each other; each has a single bond to the outside.
Answer:
C. Two hexagon rings with carbons at each corner have alternating double bonds and share one side. H is single bonded to all the C's except the ones on the shared side
Explanation:
The structure of aromatic hydrocarbon contain benzene (C6H6) which is a cyclic hydrocarbon.
Alternating single and double bond at each corner of two hexagon rings is an aromatic hydrocarbon as the alternating single and double bond form a benzene ring and H is single bonded to all the carbon's (C) except on the shared side.
Hence, the correct answer is "C"
Answer:
if you look up "propyne hydrocarbon" online, the model of D should pop up!
H - C = C - C - H
Middle C after "=" should have H on both sides (top and bottom)
Explanation:
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
If the SN2 reaction of an aromatic alcohol with an alkyl halide, like the synthesis of nerolin, is successful, what changes would be seen in the IR spectrum for the product compared to the starting material
Answer:
O-H stretch signal at 3300 cm-1
Explanation:
In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).
In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).
I hope it helps!
In movies about space, there is frequently a space battle scene where ships
explode in big fireballs. Why are these scenes unscientific?
O
A. The explosion would not be a fireball, but would point towards the closest gravity
source, a planet or star. They are only fireballs on Earth because we are already on a
gravity source.
B. None of these
C. There is no oxygen in space, so there can be no combustion.
OD. Space is very cold, there would not be enough heat energy for an explosion to occur.
-
The correct answer is C. There is no oxygen in space, so there can be no combustion.
Explanation:
Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.
This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).
Answer:
C.) There is no oxygen in space, so there can be no combustion.
Explanation:
I got it correct on founders edtell
suppose you are titrating vinegar, which is an acetic acid solution
Answer:
0.373 M
Explanation:
The balanced equation for the reaction is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, the following were obtained:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Next, we shall write out the data obtained from the question. This include:
Volume of base, NaOH (Vb) = 32.17 mL
Molarity of base, NaOH (Mb) = 0.116 M
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
The molarity of the acid solution can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.116 x 32.17 = 1
Cross multiply
Ma x 10 = 0.116 x 32.17
Divide both side by 10
Ma = (0.116 x 32.17) /10
Ma = 0.373 M
Therefore, the concentration of the acetic acid is 0.373 M.
True or False
1. Density is considered a chemical (i.e., not a physical) property. TRUE FALSE
2. When naming an ionic compound containing a transition element such as iron (Fe), the name must include a Roman numeral to indicate the charge of the metal ion. TRUE FALSE
3. The neutron was discovered about 20 years after the electron and proton because it has no charge (in order for it to be detected). TRUE FALSE
4. When we balance a chemical equation, we are observing the law of conservation of mass as well as the part of Dalton’s theory that atoms are neither created or destroyed in a chemical reaction TRUE FALSE
5. When a gas is heated up in a closed container, the kinetic energy of the molecules or atoms of the gas increase, which leads to a decrease in the pressure of the gas. TRUE FALSE
6. The amount of enthalpy (heat energy) for a reaction is directly proportional to the amount (number of moles or grams) of the reactants. TRUE FALSE
7. The combined gas law works for any gas (i.e., you do not need to know the chemical formula). TRUE FALSE
8. A balloon with 10.0 g of CO2 gas will have more molecules than a 10.0 g sample of NO gas. TRUE FALSE
9. Unless a sample is at absolute zero (kelvins), the particles in the sample will have kinetic energy and have some kind of motion. TRUE FALSE
Answer:
1. False
2. True
3. True
4. True
5. True
6. True
7. True
8. False
9. True
Explanation:
Density is a physical property since its measurement does not involve any chemical process.
Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.
Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.
The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.
When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.
Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.
The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.
10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.
If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.
AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.
Answer:
Ksp = 2.06x10⁻¹⁰
Explanation:
For AB₂. solubility product constant, Ksp, is written as follows:
AB₂(s) ⇄ A²⁺ + 2Br⁻
Ksp = [A²⁺] [Br⁻]²
Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:
AB₂(s) ⇄ A²⁺ + 2Br⁻
AB₂(s) ⇄ X + 2X
where X are moles that are soluble (Molar solubility)
Ksp = [X] [2X]²
Ksp = 4X³As molar solubility of the salt is 3.72x10⁻⁴M:
Ksp = 4X³
Ksp = 4(3.72x10⁻⁴)³
Ksp = 2.06x10⁻¹⁰A chemist adds of a M sodium carbonate solution to a reaction flask. Calculate the millimoles of sodium carbonate the chemist has added to the flask. Round your answer to significant digits.
Answer:
256 milimoles of Sodium Carbonate are in the flask
Explanation:
The chemist adds 180mL of a 1.42M sodium carbonate solution.
Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (Sodium Carbonate, in this case) per liter of solution.
A 1.42M solution contains 1.42 moles of Sodium carbonate in 1L of the solution.
As you have 180mL = 0.180L, moles of sodium carbonate you have are:
0.180L × (1.42 moles Sodium Carbonate / 1L) = 0.2556 moles of sodium carbonate
1000milimoles are equal to 1mole. 0.2556 moles are:
0.2556 moles × (1000 milimoles / 1 mole) =
256 milimoles of Sodium Carbonate are in the flaskIn the laboratory you are asked to make a 0.694 m copper(II) iodide solution using 455 grams of water. How many grams of copper(II) iodide should you add
Answer:
100.2g of CuI₂ you must add
Explanation:
Molality, m, is defined as the ratio between moles of solute and kg of solvent.
In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: 317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.
As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:
0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂
Using molar mass, grams of CuI₂ in the solution are:
0.316moles CuI₂ ₓ (317.35g / mol) =
100.2g of CuI₂ you must addwhat is the net ionic equation with its physical states? (NH4)2CO3(aq)+Ca(ClO4)2(aq)⟶CaCO3(s)+2NH4ClO4(aq)
Answer: The net ionic equation is [tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex](NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)[/tex]
The equation can be written in terms of ions as:
[tex]2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)[/tex]
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The ions which are present on both the sides of the equation are ammonium and chlorate ions and hence are not involved in net ionic equation.
Hence, the net ionic equation is :
[tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and NO3−(aq) Cr(OH)2(s) and NaNO3(s) Cr2(OH)2(aq) and NaNO3(aq) Cr(OH)2(aq) and NaNO3(s) Cr(OH)2(s), N2(g), and H2O(l)
Answer:
Cr(OH)2(s), Na+(aq), and NO3−(aq)
Explanation:
Let is consider the molecular equation;
2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)
This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.
Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.
How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 660.0 mL of a 0.0577 M succinic acid solution to produce a pH of 5.869 ? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).
Answer:
the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g
Explanation:
Given that:
The volume of [tex]K_2C_4H_4O_4.3H_2O[/tex] = 660.mL
The molarity of succinic acid = 0.0577 M
The pH of the solution = 5.869
The pKa₁ = 4.207
The pKa₂ = 5.636
The pKa₂ is required to be used for the determination of given that succinic acid is dissociated twice.
Using Henderson-HasselBalch Equation,
[tex]PH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]
we know that :
The volume of [tex]K_2C_4H_4O_4[/tex] = 660.mL
The molarity of succinic acid = 0.0577 M
number of moles = Molarity [tex]\times[/tex] Volume (liters)
number of moles = 0.0577 [tex]\times[/tex] 0.660 L
number of moles = 0.038082 mol
The balanced chemical reaction for this equation can be expressed as follows:
[tex]K_2C_4H_4O_4+C_4H_4O_4^{2-} \longleftrightarrow 2KC_4H_4O_4^-[/tex]
Here, the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]
number of moles of [tex]2KC_4H_4O_4^-[/tex] = 2 × 0.038082 mol
number of moles of [tex]2KC_4H_4O_4^-[/tex] = 0.076164 mol
From the Henderson-HasselBalch Equation,
[tex]pH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]
[tex]pH =pKa_2 + log\dfrac{[K_2C_4H_4O_4]}{[KC_4H_4O_4^-]}[/tex]
[tex]pH =pKa_2 + log\dfrac{n_{K_2C_4H_4O_4}}{n_{KC_4H_4O_4^-}}[/tex]
[tex]5.869 =5.636+ log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]0.233= log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]10^{0.233}= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]1.710015315= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]1.710015315 \times 0.076164= n_{K_2C_4H_4O_4[/tex]
[tex]n_{K_2C_4H_4O_4}= 0.1302416[/tex]
SInce; the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]
[tex]n_{C_4H_4O_4^{2-}} =[/tex]( 0.1302416 + 0.038082) mol
[tex]n_{C_4H_4O_4^{2-}} =[/tex] 0.1683236 mol
The grams of dipotassium succinate trihydrate = numbers of moles of dipotassium succinate trihydrate × Molar mass
The grams of dipotassium succinate trihydrate = 0.1683236 mol × 248.32 g/mol
The grams of dipotassium succinate trihydrate = 41.798g
Thus , the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g
The correct IUPAC name for the following compound is
Answer:
1-cyclopentylhexan-2-one
Explanation:
1-cyclopentylhexan-2-one
a) During the workup of the reaction, an aqueous solution of sodium bicarbonate was added to the cooled reaction mixture. Why was this done
Answer:
The purpose of adding an aqueous solution of sodium bicarbonate is either to extract a certain compound or to remove/neutralize acidic compounds present in the reaction mixture.
Explanation:
Upon adding sodium bicarbonate, carbon dioxide is released (gaseous state at room temperature), which helps build up pressure that is able to push out the unwanted gas/liquid.
Zn + 2 HCl --> H2 + ZnCl2 If 1.70 g of Zn are reacted, how many grams of ZnCl2 can be created? Show work and process and I will give brainliest
Explanation:
first find the the number of moles of of zinc .
as the number of moles of zinc and ZnCl2 is same we can calculate the mass of ZnCl2.
The height of a column of mercury in a closed-end manometer is 13.2 cm. What is the pressure of the gas in torr
Answer:
[tex]P_{gas}=131.96torr[/tex]
Explanation:
Helo,
In this case, the pressure must be computed as follows:
[tex]P_{gas}=gh\rho _{Hg}[/tex]
Which is using the density of mercury (13.6 g/mL) and its height, thus, we obtain (using the proper units):
[tex]P_{gas}=9.8\frac{m}{s^2}*(13.2cm *\frac{1m}{100cm} )*(13.6\frac{g}{cm^3}*\frac{1kg}{1000g}*\frac{1x10^6cm^3}{1m^3} )\\\\P_{gas}=17592.96Pa*\frac{760torr}{101325Pa} \\\\P_{gas}=131.96torr[/tex]
Best regards.
what is 1 +1 (a) 11 (b) 3 (c) 6 (d) 2
Answer:
(d) 2
Explanation:
Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer
A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
Answer:
[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]
Explanation:
There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the Al be Component 1 and the H₂O be Component 2.
Data:
For the Al:
[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the water:
[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
(a) The relative temperature changes
[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]
(b) Final temperature
[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]
Check:
[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]
The second term has only two significant figures because ΔT₂ has only two.
It agrees to two significant figures
In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.
Answer:
This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.
Explanation:
In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.
Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.
So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.
Arrange the following oxides in order of increasing acidity.
Rank from least acidic to most acidic. To rank items as equivalent,overlap them.
CaO
P2O5
SO3
SiO2
Al2O3
CO2
Answer:
Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.
With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:
CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.
Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Note the following:
Acidity of an oxide depends on its electronegativity.Non-metals are more electronegative, while metals are less electronegative.Acidity of oxides increases across a period as you move from left to the right side of a periodic table.Acidity of oxides decreases down a group (column) in a periodic table.Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.
CaO, is the least, because it is an oxide of the metal, Calcium, which is at the far left in group 2 in the periodic table.The next is, [tex]Al_2O_3[/tex]. Aluminum is a metal from group 3.[tex]SiO_2[/tex] is an oxide of Silicon, also in group 4 but below Carbon.[tex]CO_2[/tex] is an oxide of Carbon, from group 4.
[tex]P_2O_5[/tex] is an oxide of the non-metal, Phosphorus, a group 5 element
[tex]SO_3[/tex] is an oxide of the non-metal, Sulphur, a group 6 element.
Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Learn more here:
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The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.
Answer:
Mass PbCl₂ = 50.24g
Mass AgCl = 14.84g
Explanation:
The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:
Ag⁺ + Cl⁻ → AgCl(s)
Pb²⁺ + 2Cl⁻ → PbCl₂(s)
If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:
Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂
And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:
(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl
Moles of Cl⁻ that were added in the KCl solution are:
0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.
Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)
0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles
0.45409 + 0.00021X = 0.46464
0.00021X = 0.01055
X = 0.01055 / 0.00021
X = 50.24g
As X = Mass PbCl₂
Mass PbCl₂ = 50.24gAnd mass of AgCl = 65.08 - 50.24
Mass AgCl = 14.84gThe masses of the compounds in the precipitate can be found my knowing
the number of moles of chloride ion contributed by each compound.
The mass of PbCl₂ in the precipitate is approximately 49.24 gThe mass of AgCl in the precipitate is approximately 15.84 gReasons:
The given parameter are;
Volume of KCl solution added = 0.242 L
Concentration of KCl solution = 1.92 M KCl
The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions
Precipitates formed = AgCl and PbCl₂
The mass of the precipitate = 65.08 g
Required:
The mass of PbCl₂ and AgCl in the precipitate
Solution;
Number of moles of chloride ions in a mole of PbCl₂ = 2 moles
Number of moles of chloride ions in a mole of AgCl = 1 mole
Let X represent the mass of PbCl₂ in the precipitate, we have;
The mass of AgCl in the precipitate = 65.08 g - X
[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]
Number of moles of chloride ions from PbCl₂ is therefore;
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]
The number of moles of chloride ions from one mole of KCl = 1 mole
Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;
0.242 L × 1.92 moles/L = 0.46464 moles
Number of moles of chloride ions from KCl = 0.46464 moles
[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]
Which gives;
[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]
Therefore;
[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]
The mass of PbCl₂ in the precipitate, X ≈ 49.24 g
The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g
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A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture
Answer:
0.641 g of Nitrogen are present in the mixture.
Explanation:
We use the Ideal Gases Law, to solve this question.
For the mixture:
P mixture . V mixture = mol mixture . R . T
We convert the T° to K → 23°C + 273 = 296 K
R = Ideal gases constant → 0.082 L.atm/mol.K
1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K
2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles
We know that sum of partial pressure = 1
Partial pressure N₂ + Partial pressure O₂ = 1
1 - 0.722 atm = Partial pressure N₂ → 0.278 atm
We apply the mole fraction concept:
Partial pressure N₂ / Total pressure = Moles N₂ / Total moles
Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles
Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles
We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g
641 mg
A cannonball made of iron has a volume of 121 cm3. What is the mass, in kilograms, of the cannonball? Express your answer to three significant figures and include the appropriate units.
Answer:the mass is 35456
Explanation:
because it works
The density has been the division of the mass to the volume of the substance. The mass of an iron cannonball of density 7.86 grams/cm³ is estimated to be 0.951 kilograms.
What is the relationship between mass and density?The mass (m) has been calculated by multiplying the density (D or ρ) and volume (V) of the iron cannonball. The measurement of the amount is called mass whereas the measurement of the space occupied by the substance volume is given as density.
The amount of the substance contained in a volume (milliliters, liters, and cubic centimeters) along with the density of the substance has been used to give mass. The mass can be estimated in kilograms, grams, milligrams, etc.
Given,
Density of iron cannonball (D) = 7.86 grams/cm³
Volume of cannonball (V) = 121 cm³
The formula to calculate mass is given as:
Density = mass ÷ volume
Mass = density × volume
Substituting values in the above formula, the mass of the iron cannonball in grams is calculated as:
mass = 7.86 grams/cm³ × 121 cm³
mass = 951.06 grams
Therefore, the iron cannonball with a density of 7.86 grams/cm³ has a mass of 0.951 kilograms.
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Your question is incomplete, but most probably your full question was, A cannonball made of iron has a volume of 121 cm³. if the iron has a density of 7.86 g/cm³, what is the mass, in kilograms, of the cannonball?
Determine the cell notation for the redox reaction given below.
Sn(s) + 2H+(aq) ⟶ Sn2+(aq) + H2(g)
a. H+(aq) | H2(g) | Pt ∥ Sn(s) | Sn2+(aq)
b. H2(g) | H+(aq) | Pt ∥ Sn2+(aq) | Sn(s)
c. Sn2+(aq) | Sn(s) ∥ H2(g) | H+(aq) | Pt
d. Sn(s) | Sn2+(aq) ∥ H+(aq) | H2(g) | Pt
e. Sn(s) | H2(g) ∥ Sn2+(aq) | H+(aq) | Pt
Answer:
The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt
Explanation:
The half reactions are:
2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)
Sn(s) ⟶ Sn²⁺(aq) + 2 e- (oxidation)
In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:
anode reaction∥ cathode reaction
where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.
In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).
The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:
Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt
A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.
Answer:
0.60 mol
Explanation:
There is some info missing. I think this is the original question.
Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.
Step 1: Given data
Moles of water required: 1.5 mol
Step 2: Write the balanced equation
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water
The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol