Answer:
Rate constants are temperature dependent.
Explanation:
Reaction rate is used to quantify the rate of chemical reaction. There is a relationship between the reaction rate and the half-life of the reaction and the Gibbs free energy of activation, and the reaction rate is temperature dependent according to the equation.
For a reaction shown below
a A + b B ⇒ c C
The rate of reaction of the reaction is given by
[tex]r = k(T) [A]^{m}[B]^{n}[/tex]
where k(T) is the reaction constant, which is seen to be dependent on the temperature of the reaction.
Also, k(T) is numerically equal to
[tex]k(T) = Ae^{\frac{E_{a} }{RT} }[/tex]
where
r = reaction rate
A = pre exponential factor
[tex]E_{a}[/tex] = Activation energy
R = gas constant
T = temperature
and m and n are experimentally determined partial orders in [A] and [B]
A radioactive isotope of mercury, 197Hg, decays togold, 197Au, with a disintegration constant of 0.0108 h-1. What fraction of a sample will remain at the end of three half-lives (
Answer:
THE FRACTION OF THE SAMPLE REMAINING AFTER THREE HALF LIVES IS 0.125 OR 125/1000
Explanation:
A radioactive isotope of mercury decay to gold with a disintegration constant of 0.0108 h^-1
To calculate the fraction of sample remaining after three half life, we first calculate the half life of the decay.
Half life = ln 2 / Y
Y = disintegration constant
So therefore,
half life = ln 2 / 0.0108
half life = 0.693 / 0.0108
half life = 64.18 hours.
So a decay occurs after 64.18 hours.
To calculate the fraction remaining after 3 half life:
N(t) = N(o) e ^-Yt
where t = 3 half life
So, N / No = e^-Y ( 3 t1/2)
Since t 1/2 = ln 2 / Y, so we can re-write the formula as:
Nt / No = e^-Y ( 3 ln 2/ Y)
Nt / No = e^-3 ln2
Nt / No = e^-3 * 0.693
Nt / No = e^-2.079
Nt / No = 0.125
So the fraction of the sample remaining after 3 half lives is 125/ 1000 or 0.125
What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.
Answer: B. The period number tells which is the highest energy level occupied by the electrons
Explanation:
The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.
hence, the correct option is B. The period number tells which is the highest energy level occupied by the electrons.
The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.
What are energy levels ?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.
When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.
Therefore, option B , second option is correct.
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How many atoms of hydrogens are found in 3.21 mol of
C3H8?
Answer:
1.55 × 10²⁵ atoms of H
Explanation:
3.21mol C₃H₈ × 8mol H × (6.022×10²³)
Why is it important that the primary standard chemical be non-hygroscopic and pure? Why is it important to dry the primary standard to a constant weight?
Answer:
It is extremely important for the primary standard chemical to be non – hygroscopic and pure and to also have a constant weight because you don't want any moisture or any impurities to alter the stoichiometric point in the reaction
It is important that the primary standard chemical be non-hygroscopic and pure to calculate the exact calculation of the reaction.
What is non hygroscopic chemicals?Non hygroscopic chemicals are those compounds which will not absorb water or mositure from the outside.
If we take any substance which are hygroscopic in nature and during the chemical reaction if they absorb water content or moisture then the mass of that substance will alter and changes all the calculation of the reaction.
So, to maintain the stability of calculation we use non hygroscopic materials.
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2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations
Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
The blending of one s atomic orbital and three p atomic orbitals produces ________.
A three sp3
B four sp3
C three sp
D four sp2
E four sp
Answer:
B. four sp3
Hope that helps.
We have that for the Question "The blending of one s atomic orbital and three p atomic orbitals produces?"
Answer:
Option B = four [tex]sp^3[/tex]
Explanation:
When 1 s orbital blends with 3 p orbitals, they form a tetrahedrical shaped figure with each being a [tex]SP^3[/tex] orbital.. A total of 4 orbitals
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Why was it important to establish the Clean Air Act?
Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.
Explanation:
Answer:
Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.
How many Liters of 0.968M solution can be made if 0.581 moles of solute are added? Group of answer choices 0.600 L 60 mL 0.562 L 1.00 L
Answer:
0.6L
Explanation:
The formula of molarity is molSolute/litreSolution
[tex]0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L[/tex].
1. Define the Law of Conservation of Mass (via text). Now that you’ve defined this law, explain what it means in your own words using an example.
Explanation:
The law of conservation of mass states that mass can neither be created nor be destroyed.
Explanation in own words = this means that in this universe no one can create or destroy mass.
No physical or chemical force.
Write the electron configuration when Sulfur gains two electrons
Answer:
Explanation:
If sulfur gains 2 electrons then two electrons should be added to it electronic configuration.
The decomposition of hydrogen peroxide to form water and oxygen gas releases 196.6 kJ per mole of hydrogen peroxide. This reaction occurs when hydrogen peroxide is placed on a cut to sterilize it. How much heat is released when 5.70 mol H2O2 decompose
Answer:
1120.62 kJ
Explanation:
In order to find how much heat is released for 7.70 mol, we have to compare it with the heat released from one mole.
So from the question, we have;
196.6 kJ = 1 mol
x = 5.70
x = 5.70 * 196.6 / 1
x = 1120.62 kJ
230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen
Answer:
15.3 %
Explanation:
Step 1: Given data
Mass of the sample (ms): 230 gMass of carbon (mC); 136.6 gMass of hydrogen (mH): 26.4 gMass of nitrogen (mN): 31.8 gStep 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Choose the most appropriate indicator for the titration of a weak acid with NaOH, where the expected equivalence point of the titration is at pH 8.8.
a. methyl orange, pH range 3.2-4.4
b. methyl red, pH range 4.8 6.0
c. bromothymol blue, pH range 6.0-7.6
d. phenolphthalein, pH range 8.2-10.0
e. alizarin yellow R. pH range 10.1-12.0
Answer:
D phenolphthalein,pH range 8.2-10.0
Calculate the concentration of H3O+ in a solution that contains 6.25 × 10-9 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Answer:
[OH⁻] = 1.60 × 10⁻⁶ M
Basic
Explanation:
Step 1: Given data
Concentration of H₃O⁺: 6.25 × 10⁻⁹ M
Step 2: Calculate the concentration of OH⁻
We will use the following expression.
Kw = [H₃O⁺] × [OH⁻] = 1.00 × 10⁻¹⁴
[OH⁻] = 1.00 × 10⁻¹⁴ / 6.25 × 10⁻⁹ = 1.60 × 10⁻⁶ M
Step 3: Calculate the pH
We will use the following expresion.
pH = -log [H₃O⁺] = -log (6.25 × 10⁻⁹) = 8.20
Since the pH > 7, the solution is basic.
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached to the positive terminal of a potentiometer and the calomel electrode was attached to the negative terminal.(a) Write a half-reaction for the Cu electrode. (Use the lowest possible coefficients. Omit states-of-matter.)
(c) Calculate the cell voltage.
Answer:
(a) Cu²⁺ +2e⁻ ⇌ Cu
(c) 0.07 V
Explanation:
(a) Cu half-reaction
Cu²⁺ + 2e⁻ ⇌ Cu
(c) Cell voltage
The standard reduction potentials for the half-reactions are+
E°/V
Cu²⁺ + 2e⁻ ⇌ Cu; 0.34
Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241
The equation for the cell reaction is
E°/V
Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34
2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; -0.241
Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10
The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation
(ii) Calculations:
T = 25 + 273.15 = 298.15 K
[tex]Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}[/tex]
Heating carvone with aqueous sulfuric acid converts it into carvacrol. The mechanism involves the following steps:
1. The terminal alkene of carvone reacts with acid to form tertiary carbocation 1;
2. A hydride shift results in the formation of tertiary carbocation 2;
3. Deprotonation of the ring leads to conjugated diene 3;
4. Deprotonation at the α carbon leads to the product carvacrol.
Required:
Draw the mechanism and then draw the structure of tertiary carbocation 2.
Answer:
See figure 1
Explanation:
In this question, we have to start with the protonation of the double bond. In carvone we have two double bonds, so, we have to decide first which one would be protonated.
The problem states that the terminal alkene is the one that would is protonated. Therefore, we have to do the protonation in the double bond at the bottom to produce the carbocation number 1. Then, a hydride shift takes place to produce the carbocation number 2. A continuation, an elimination reaction takes place to produce the conjugated diene. Then the diene is protonated at the carbonyl group and with an elimination reaction of an hydrogen in the alpha carbon we can obtain carvacol.
Molarity of NaOH: From the following data calculate molarity of NaOH. Molar mass of KHP is 204.23 g/mol. Show calculation. Mass of Erlenmeyer flask + KHP 84.847 g Mass of Erlenmeyer flask 84.347 g Mass of KHP ??? Final buret reading 12.25 mL Initial buret reading 0.50 mL Volume of NaOH added ???
Answer:
Explanation:
Mass of Erlenmeyer flask + KHP = 84.847 g
Mass of Erlenmeyer flask = 84.347 g
Mass of KHP = .5 g
moles of KHP = .5 / 204.23
= 2.448 x 10⁻³ moles
moles of NaOH reacted = 2.448 x 10⁻³
Final buret reading = 12.25 mL
Initial buret reading = 0.50 mL
Volume of NaOH added=
4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50 L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O 2(g) 2NO 2(g) is reached when [NO] = 1.6 M. Calculate K c for the reaction at this temperature.
Explanation:
At 593K a particular decomposition’s rate constant had a value of 5.21×10−4 and at 673K the same reaction’s rate constant was 7.42×10−3. It was noticed that when the reactant’s initial concentration was 0.2264 M (with a 593K reaction temperature), the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Recall that rate laws have the form rate = k [A]x and, showing work, determine the order of the decomposition reaction.
14. Based on your previous observations, predict the impact of changing the number of moles of a gas sample on the volume of the gas sample (if pressure and temperature are held constant). What effect would changing the number of moles of a gas sample have on the temperature of a gas sample (if pressure and volume are held constant)? Explain
Answer:
Number of moles of gas is directly proportional to the volume of the gas
Number of moles of the gas is directly proportional to the temperature of the gas
Explanation:
According to Avogadro's law, changing the number of moles of a gas changing the volume of the gas also since the volume of a gas is directly proportional to the number of moles of the gas.
Hence from Avogadro's law; V= kn where k is a proportionality constant, V is the volume of the gas and n is the number of moles of the gas.
Changing the number of moles will also lead to a change in the temperature of the gas, since volume is directly proportional to the number of moles of the gas and volume is also directly proportional to temperature (Charles law), it the follows that number of moles of the gas is directly proportional to its temperature.
How many grams are in 5.87 x 10^21 molecules of sulfur?
Answer:
0.312g
Explanation:
From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules
1mole of sulphur = 32g
If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules
Then, Xg of sulphur will contain 5.87x10^21 molecules i.e
Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g
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Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Answer:
The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.
Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
Considering the definition of ionization energy,
Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.
You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.
In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.
In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.
Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
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https://brainly.com/question/24409114https://brainly.com/question/14158485?referrer=searchResultshttps://brainly.com/question/14454446?referrer=searchResults2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = –118 kJ Calculate the heat when 250.0 mL of 0.500 M HCl is mixed 500.0 mL of 0.500 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0 oC and that the final mixture has mass of 750.0 g and a specific heat capacity of 4.18 J oC–1g–1, calculate the final temperature (in oC) of the mixture.
Answer:
Heat = 7375J
Final temperature of the mixture = 27.35°C
Explanation:
In the reaction:
2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l) ΔH = –118 kJ
When 2 moles of HCl reacts with excess of Ba(OH)₂ there are released 118kJ.
In the reaction, moles of HCl and Ba(OH)₂ that reacts are:
Moles HCl = 0.250L ₓ (0.500 moles / L) = 0.125 moles HCl
Moles Ba(OH)₂ = 0.500L ₓ (0.500 moles / L) = 0.250 moles Ba(OH)₂
For a complete reaction of 0.125 moles of HCl you need:
0.125 mol HCl ₓ (1 mole Ba(OH)₂ / 2 moles HCl) = 0.0625 moles Ba(OH)₂
As you have 0.250 moles of Ba(OH)₂, this reactant is in excess
2 moles of HCl that react release 118kJ, 0.125 moles of HCl release:
0.125 moles HCl ₓ (118kJ / 2 moles) = 7.375kJ =
7375JThe heat released can be obtained with the formula:
Q = C×m×ΔT
Where Q is heat, C specific heat of the solution, m its mass and ΔT change in temperature.
Replacing:
Q = C×m×ΔT
7375J = 4.18J/g°C×750.0g×ΔT
2.35°C = ΔT
As ΔT = Final T - Initial T:
2.35°C = Final T - 25.0°C
27.35°C = Final temperature of the mixture
The adiabatic saturation and wet-bulb temperatures will be equivalent for atmospheric air when the two are approximately equal at atmospheric temperatures and pressure.
a. True
b. False
Copper was one of the earliest metals used by humans, because it can be prepared from a wide variety of copper minerals, such as cuprite (Cu2O), chalcocite (Cu2S), and malachite [Cu2CO3(OH)2]. Balance the following reactions for converting these minerals into copper metal. Place a coefficient in each gray box.
(a) Cu2O(s) + C(s) rightarrow Cu(s) + CO2(g)
(b) Cu2O(s) + Cu2 S(s) rightarrow Cu(s) + SO2(g)
(c) Cu2 CO3 (OH)2(s) rightarrow CuO(s) + CO2(g) + H2O(g)
Use the left and rightarrow keys to move the cursor out of a superscript or subscript in the module.
Answer:
a. 2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)
b. 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
c. Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
Explanation:
A reaction is balanced when you have the same amount of atoms in reactants and products.
In the reactions:
(a) Cu₂O(s) + C(s) → Cu(s) + CO₂(g)
As a general rule, you first balance oxygen and hydrogen. In products you have 2 oxygens, then:
2 Cu₂O(s) + C(s) → Cu(s) + CO₂(g)
Carbon is balanced yet. Thus, you need just to balance Cu:
2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)
(b) Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)
Balancing oxygen:
2Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)
Sulfur is balanced yet. Now you just need to balance Cu:
2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
(c) Cu₂CO₃(OH)₂(s) → CuO(s) + CO₂(g) + H₂O(g)
This reaction is different because the reactant is a chemical with a lot of atoms. we will first balance Cu:
Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
Balancing copper, oxygen, hydrogen and carbon are balanced:
Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
Which of these substances has the highest pOH? 0.10 M HCl, pH = 1 0.001 M HNO3, pH = 3 0.01 M NaOH, pH = 12 The answer is 0.10 M HCI, pH=1
Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.
Explanation:
The 0.10M HCI, pH = 1 solution has the highest pOH. Therefore, option (1) is correct.
What is the pOH?pOH of a solution can be determined from the negative logarithm of the hydroxide ions concentration in the solution.
The mathematically pOH of the solution can be expressed as:
pOH = -log [OH⁻] ..............(1)
Where [OH⁻] represents the concentration of hydroxide ions in an aqueous solution.
Given, the pH = 1 of HCl
pH + pOH = 14
1 + pOH = 14
pOH = 14 - 1
pOH = 13
Given, the pH = 3 of HNO₃
pH + pOH = 14
3 + pOH = 14
pOH = 14 - 3
pOH = 11
Given, the pH = 12 of NaOH = 0.01 M
pH + pOH = 14
12 + pOH = 14
pOH = 14 - 12
pOH = 2
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When two molecules of methanol (CH3OH) react with oxygen, they combine with three O2 molecules to form two CO2 molecules and four H2O molecules. How many H2O molecules are formed when 94 methanol molecules react
Answer:
188
Explanation:
For every 2 molecules of methanol reacted, 4 molecules of water are formed. Use this relationship to solve.
2/4 = 94/x
2x = 376
x = 188
188 molecules of water will be formed.
Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
what are the similarities between amorphous solid and crystalline solid
Answer:
solid dont know
Explanation:
so sorry ask another
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.
What is volume?Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.
Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.
This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.
Here, it is given that mass of olive oil is 1.2kg.
We know that,
Density of olive oil = 0.917kg/l.
Volume = mass/density
Volume = 1.2/0.917.
Volume = 1.30 lit.
Volume = 1300mL.
Thus, the volume of olive oil will be 1300 mL.
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what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equation
Answer:
[tex]Cu~+~PtCl_2->Pt~+~CuCl_2[/tex]
Explanation:
In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case [tex]Pt^+^2[/tex]) and the anion is the atom at the right of the name (in this case [tex]Cl^-[/tex]). With this in mind, the formula would be [tex]PtCl_2[/tex].
Now, if we used metallic copper we have to put in the reaction only the copper atom symbol [tex]Cu[/tex]. So, we have as reagents:
[tex]Cu~+~PtCl_2->[/tex]
The question now is: What would be the products? To answer this, we have to remember "single displacement reactions". With a general reaction:
[tex]A~+~BC->AB~+~C[/tex]
With this in mind, the reaction would be:
[tex]Cu~+~PtCl_2->Pt~+~CuCl_2[/tex]
I hope it helps!