Answer:
Oxygen is oxidized and hydrogen is reduced
Explanation:
In the electrolysis of water, a pair of Platinum electrodes are immersed in water. Th water has a small quantity of either an acid, salt or base, in most cases H2S04, added to it, to aid ionization. This is because water on its own does not posses enough ions to undergo electrolysis. At the platinum anode, water is oxidized to oxygen gas and hydrogen ions. At the platinum cathode, water is reduced to hydrogen gas and hydroxide ions. The proportion of oxygen and hydrogen produced should be theoretically 1 : 2 respectively, but is not usually so, due to competing side reactions. The hydrogen by product is usually used as a fuel source, and it usually combine with the hydroxide ion to form water back again.
Consider the following reaction: Br2(g) + 3 F2(g) LaTeX: \rightarrow→ 2 BrF3(g) LaTeX: \Delta H_{rxn}Δ H r x n= ‒836 kJ/mol Bond Bond Energy (kJ/mol) Br–Br 193 F–F 155 Using the above bond dissociation energies, calculate the energy, in kJ/mol, of a Br–F bond.
Answer: The energy of a Br–F bond is 110 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]
[tex]B.E_{Br-F}=110kJ/mol[/tex]
Thus the energy, in kJ/mol, of a Br–F bond is 110
What is the term used by particular kind of matter Called??
Answer:
[tex]\boxed{\mathrm{substance}}[/tex]
Explanation:
The term used by particular kind of matter is called substance.
A substance is a particular kind of matter because it has physical properties.
Give the formulas for all of the elements that exist as diatomic molecules under normal conditions. See if you can do this without looking anything up.
Answer:
They are:
H2, N2, O2, F2, Cl2, Br2, and I2.
Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.
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The charcoal from ashes found in a cave gave 7.4 14C counts per gram per minute. Wood from the outer portion of a growing tree gives a comparable count of 15.3. The half-life of 14C is 5700 years.
How old are the ashes?
A) 3245 y
B) 5700 y
C) 5970 y
D) 9220 y
E) 5437 y
Answer:
C) 5970 y
Explanation:
Given;
initial amount of wood, N₀ = 15.3 cpm/g
remaining amount of wood (charcoal), N = 7.4 cpm/g
half life of carbon 14, t 1/2 = 5700 years
The age of the ashes can be calculated using the following formula;
[tex]N = N_0(\frac{1}{2})^{\frac{t}{t_1_/_2} }\\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = \frac{N}{N_0} \\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = \frac{7.4}{15.3} \\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = 0.48366\\\\t = t_{1/2} Log\frac{1}{2} (0.48366)\\\\t = \frac{t_{1/2}ln(0.48366)}{-ln(2)} \\\\t = t_{1/2}(1.0479)\\\\t = 5700(1.0479)\\\\t = 5973 \ years\\\\t = 5970 \ years(nearest \ ten)[/tex]
Therefore, the ashes are 5970 years
19. Hexavalent chromium bonds with fluorine to form an ionic compound. What's the chemical formula and name for this compound
using the Stock system?
A. Cr2F6, chromous hexafluoride
B. CrF6, chromic fluoride
C. CrF6, chromium(VI) hexafluoride
D. CrF6, chromium(VI) hexafluoride
Answer:
C. CrF6, chromium(VI) hexafluoride.
Explanation:
Hello,
In this case, since we are given a hexavalent chromium we must notice it has +6 as its oxidation state. Moreover, fluorine, when forming ionic compounds works with -1, for which the chemical formula is:
[tex]Cr^{6+}F^-\\\\CrF_6[/tex]
And the stock name is indeed C. CrF6, chromium(VI) hexafluoride (looks like D. is the same) since we have six fluoride ions in the formula and we point out chrmium's oxidation state.
Regards.
Answer:
C. CrF6, chromium(VI) hexafluoride.
Explanation:
Natural atom of the same element may have the same _________?
A)proton
B)neutron
C)electron
D)All
Answer:B
Explanation:
Answer: i think it is c
Explanation: i checked my textbook.
A study of the following system, 4 NH3(g) + 7 O2(g) <--> 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [N2O4] = [H2O] = 3.60 M as the only components initially. At equilibrium, [H2O] is 0.600 M. Calculate the equilibrium concentration of O2(g).
Answer:
3.50 M
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
Step 2: Make an ICE chart
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
I 0 0 3.60 3.60
C +4x +7x -2x -6x
E 4x 7x 3.60-2x 3.60-6x
Step 3: Calculate the value of x
The concentration of water at equilibrium is 0.600 M. Then,
3.60-6x = 0.600 M
x = 0.500 M
Step 4: Calculate the concentration of O₂ at equilibrium
The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M
A compound X has a molecular ion peak in its mass spectrum at m/z 136. What information does this tell us about X
Explanation:
The mass to charge ratio =136
Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X rays, ultraviolet light.
Rank from shortest to longest. To rank items as equivalent, overlap them.
infrared
green light
red light
radio waves
X rays
ultraviolet
Answer:
In the other of increasing wavelength we arrange as
X-rays
ultraviolet
green light
red light
infrared
radio waves
Explanation:
In the electromagnetic spectrum, wavelength decreases with increase in the energy of the electromagnetic wave. Since the e-m wave spectrum is arranged in the order of increasing energy (decreasing wavelength) as: Radio wave; infrared; visible light; ultraviolet; x-rays; gamma rays. Within the visible light, the green light has more energy than the red light. Therefore, the arrangement should be in the reverse direction of their increasing energy.
You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 30.0 N. You carefully add 1.25×10^4 J of heat energy to the sample and find that its temperature rises 15.0 °C. What is the sample's specific heat?
Answer:
272.33 J/Kg°C
Explanation:
Data obtained from the question include the following:
Weight of metal = 30 N
Heat used (Q) = 1.25×10⁴ J
Change in temperature (ΔT) = 15.0 °C.
Specific heat capacity (C) =..?
Next, we shall determine the mass of the metal.
The mass of the metal can be obtained as follow:
Weight (W) = mass (m) x acceleration due to gravity (g)
W = mg
Weight of metal = 30 N
Acceleration due to gravity = 9.8 m/s²
Mass (m) =..?
W = mg
30 = m x 9.8
Divide both side by 9.8
m = 30/9.8
m = 3.06 Kg
Finally, we shall determine the specific heat capacity of the metal as show below:
Heat used (Q) = 1.25×10⁴ J
Change in temperature (ΔT) = 15.0 °C.
Mass (m) = 3.06 Kg
Specific heat capacity (C) =..?
Q = mCΔT
1.25×10⁴ = 3.06 x C x 15
Divide both side by 3.06 x 15
C = (1.25×10⁴) / (3.06 x 15)
C = 272.33 J/Kg°C
Therefore, the specific heat capacity of metal is 272.33 J/Kg°C.
Based on their molecular structure, identify the stronger acid from each pair of oxyacids. Match the words in the left column to the appropriate blanks in the sentences on the right.
1) HI is a stronger acid than H2Te because iodine____than tellurium.
2) H2Te is a stronger acid than H2S because the H-Te bond is_____.
3) NaH is not acidic because hydrogen____than sodium.
a. has a more negative electron afflity
b. is more electronegative
c. has a larger atomic radius
d. stronger
e. is harder to ionize
Answer:
1)is more electronegative
2)
3) is more electronegative
Explanation:
1) for the first question, iodine is more electronegative than tellurium hence we naturally expect that HI should be more acidic than H2Te since electronegativities play a role in the acidity of chemical species.
2) the correct option is not listed because the H2Te bond is weaker than the H2S bond. This makes it easier for H2Te to dissociate releasing H^+ , thereby being more acidic than H2S.
3) Hydrogen is more electronegative than sodium hence it cannot be ionized thus NaH is not acidic.
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). What is the final temperature if 305 J of heat is added to 52.4 g of this metal, initially at 20.0 ∘C?
Answer:
65.47∘C
Explanation:
Specific heat capacity, c = 0.128 J/(g⋅∘C)
Initial temperature = 20.0 ∘C
Final temperature = ?
Mass = 52.4 g
Heat = 305 J
All these variables are related by the following equation;
H = m c ΔT
ΔT = H / mc
ΔT = 305 / (52.4 * 0.128)
ΔT = 45.47∘C
ΔT = Final Temperature - Initial Temperature
Final temperature = ΔT + Initial temperature
Final temperature = 45.47∘C + 20.0 ∘C = 65.47∘C
1. In general chemistry you learned that oxidation meant any reaction that involves loss of electrons, but in organic chemistry oxidation often refers to an earlier, more literal, definition of the terms in this experiment. What is this earlier, classical definition of oxidation
Answer:
In organic chemistry oxidation is the gain of oxygen atoms by any carbon atom or molecules.
Explanation:
Oxidation can be defined in multiple ways
1- Loss of electron
2- Increase in oxidation number
3- Loss of hydrogen
4- Gain of oxygen atoms
The last definition is the earlier one in organic chemistry.
For consultation in chemistry. whtsapp +923554232104
In reaction NH3 →3H2 + N2, how many moles of N2 formed if 2.81 g NH3 dissociate? Show work!
Answer:
0.0826 mol (corrected to 3 sig. fig.)
Explanation:
First, balance the equation:
2NH3 →3H2 + N2
Take the atomic no. of N=14.0, and H=1.0,
no. of moles = mass / molar mass
So, no. of moles of NH3 dissociated = 2.81 / (14.0+1.0x3)
= 0.165294117mol
From the equation, the mole ratio of NH3:N2 = 2:1, meaning for every 2 moles of NH3 dissociated, one mole of N2 is formed.
So, using this ratio, the no. of moles of N2 formed will be 0.165294117 / 2
=0.0826 mol (corrected to 3 sig. fig.)
What is the freezing point of a solution prepared from 45.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? Kf of water is 1.86°C/m.
Answer:
[tex]T_{sol}=-15.9\°C[/tex]
Explanation:
Hello,
In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:
[tex]T_{sol}=T_{water}-i*m*Kf[/tex]
Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:
[tex]m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m[/tex]
Thus, we obtain:
[tex]T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C[/tex]
Best regards.
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
What is freezing point depression?Freezing point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.
Step 1: Calculate the molality of the solution.We will use the definition of molality.
b = mass of solute / molar mass of solute × kg of solvent
b = 45.0 g / 62.07 g/mol × 0.0850 kg = 8.53 m
Step 2: Calculate the freezing point depression (ΔT).We will use the following expression, where Kf is the cryoscopic constant of water.
ΔT = Kf × b = 1.86 °C/m × 8.53 m = 15.9 °C
Step 3: Calculate the freezing point of the solution.The freezing point of pure water is 0°C.
T = 0°C - 15.9 °C = -15.9 °C
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
Learn more about freezing point depression here: https://brainly.com/question/14115775
Fill in the blanks
1. the ____ atoms of the ____ groups and the hydrogen atoms attached to the ____ atoms form ____.
(answer choices; nitrogen, carbon, a-helices or b-pleated sheets, carbonyl, globular of firbrous structures, oxygen, amino, random coil, thiol)
2. In the a helix, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in _____. In the B-pleasted sheet, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in ____.
(answer choices; parallell sections of a long polypeptide chain, the next amino acid unit, carboxyl, the next turn of the helical chain, thiol, amino, carbonyl)
Answer:
1. oxygen; carbonyl; nitrogen; α-helices or β-pleated sheets
2. carbonyl; amino; the next turn of the helical chain; carbonyl; amino; parallel sections of a long polypeptide chain.
Explanation:
The secondary structure of proteins is of two major conformations, the α helix and β conformations which are very stable.
1. The oxygen atoms of the carbonyl groups and the hydrogen atoms attached to the nitrogen atoms form α-helices or β-pleated sheets.
2. In the α-helix, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in the next turn of the helical chain. In the β-pleated sheet, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in parallel sections of a long polypeptide chain.
The gas in a 250. mL piston experiences a change in pressure from 1.00 atm to 2.55 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant?
Answer:
[tex]\large \boxed{\text{0.980 L}}[/tex]
Explanation:
The temperature and amount of gas are constant, so we can use Boyle’s Law.
[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]
Data:
[tex]\begin{array}{rcrrcl}p_{1}& =& \text{1.00 atm}\qquad & V_{1} &= & \text{250. mL} \\p_{2}& =& \text{2.55 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]
Calculations:
[tex]\begin{array}{rcl}\text{1.00 atm} \times \text{250. mL} & =& \text{2.55 atm} \times V_{2}\\\text{250. mL} & = & 2.55V_{2}\\V_{2} & = &\dfrac{\text{250. mL}}{2.55}\\\\& = &\textbf{98.0 mL}\\\end{array}\\\text{The balloon's new volume is $ \large \boxed{\textbf{0.980 L}}$}[/tex]
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment
Answer:
% yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
Explanation:
given
volume of acetone= 43.8 mL
molar weight of acetone = 58.08 g/mol
density of acetone = 0.791 g/mL
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)
43.8 mL = 43.8mL × 0.791g/mL
= 34.6458g ≈34.65g
1 mole of acetone = 58.08g
∴34.65g = 34.65g/58.08g
= 0.60mol
molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol
mole = mass/ molar weight
mole = 79.4g/ 234.29g/mol
mole(n) = 0.3389mol ≈ 0.34mol
1 mole of acetone will produce 1 mole of the product
∴0.60mol of acetone will produce 0.60mol of the product
but we get 0.34mol of the product
∴ % yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
The migration of atoms or molecules through a material is called Choose one: biomineralization. precipitation from a gas. solidification of a melt. diffusion.
Answer:
diffusion
Explanation:
Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration in response to a concentration gradient. A concentration gradient simply means a difference in concentration.
Diffusion occurs in solids,liquids and gases. Diffusion is fastest in gases and slowest in solids. Diffusion of solid particles may take very many years while diffusion of gases takes a few milliseconds depending on the mass of the gas.
In materials, atoms and molecules also move from one part of the material to another. This is also refereed to as diffusion.
what is chemical equation of Braium chloride?
Answer:
BaCl2
Explanation:
Barium = Ba
Chloride => Cl-
Chemical Equation:
Ba + Cl => BaCl2
Note:
The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).
What is the pressure, in Pa, if the height of a column of mercury is 400. mm and the density of mercury is 13.6 gcm3
Answer:
Explanation:
Pressure due to a liquid column
P = hdg where h is height of column , d is density of liquid and g acceleration due to gravity .
( 13.6 g cm⁻³ = 13.6 x 10³ kg m⁻³ , because 1 m³ = 10⁶ cm³ )
Putting the given values in the equation
P = 400 x 10⁻³ x 13.6 x 10³ x 9.8 Pa
= 53312 Pa
At what temperature is the following reaction feasible: Al2O3(s) + 3C(s) -> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol–1 Entropy (S) = +614 J K–1 mol–1 A. 2096.1 K B. 1273.8 K C. 477.1 K D. 1901.0 K
Answer:
Option A. 2096.1 K
Explanation:
The following data were obtained from the question:
Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹
Entropy (S) = +614 JK¯¹ mol¯¹
Temperature (T) =...?
Entropy, enthalphy and temperature are related by the following equation:
Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)
ΔS = ΔH/T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =...?
ΔS = ΔH/T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K.
Assuming an efficiency of 34.90%, calculate the actual yield of magnesium nitrate formed from 139.6 g of magnesium and excess copper(II) nitrate.Mg+Cu(NO3)2⟶Mg(NO3)2+Cu
Answer:
300.44 g
Explanation:
The balanced equation for the reaction is given below:
Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu
Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]
= 24 + 2[ 14 + 48]
= 24 + 124 = 148 g/mol
Mass of Mg(NO3)2 from the balanced equation =
1 x 148 = 148 g
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Next, we shall determine the theoretical yield of Mg(NO3)2.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2
Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g
Finally, we shall determine the actual yield of Mg(NO3)2 as follow:
Theoretical of Mg(NO3)2 = 860.87 g
Percentage yield = 34.90%
Actual yield of Mg(NO3)2 =?
Percentage yield = Actual yield /Theoretical yield x 100
34.90% = Actual yield /860.87
Cross multiply
Actual yield = 34.90% x 860.87
Actual yield = 34.9/100 x 860.87
Actual yield = 300.44 g
Therefore, the actual yield of Mg(NO3)2 is 300.44 g
A student mixes 2.83 mL of benzoyl chloride with excess 15 M NH4OH to produce 1.95 g of benzamide. What is the percent yield of this student's experiment
Answer:
Explanation:
The reaction of benzoyl chloride with NH₄OH to produce benzamide is:
Benzoyl chloride + ammonia → Benzamide + NH₄Cl
Molar mass of benzoyl chloride: 140.57 g/mol. Density 1.21g/mL
Molar mass benzamide: 121.14g/mol.
To know percent yield you must know the theoretical yield of the reaction (How many grams are produced assuming a yield of 100%). Percent yield will be (Actual yield / Theoretical Yield) ₓ 100
Moles of 2.83mL of benzoyl chloride are:
2.83mL ₓ (1.21g/mL) ₓ (1mol / 140.57g) = 0.02436 moles of benzoyl chloride.
As 1 mole of benzoyl chloride produce 1 mole of benzamide (Theoretical yield), theoretical moles of benzamide produced are 0.02436. In mass:
0.02436 moles ₓ (121.14g / mol) = 2.95g of benzoyl chloride
As there are produced just 1.95, percent yield is:
(1.95g / 2.95g) ₓ 100 = 66.1%
Cyclohexane (C6H12) undergoes a molecular rearrangement in the presence of AlCl3 to form methylcyclopentane (CH3C5H9) according to the equation: C6H12 ⇌ CH3C5H9 If Kc = 0.143 at 25°C for this reaction predict the direction in which the system
Answer:
The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C
Explanation:
Hello,
In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:
[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]
Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.
Best regards.
At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?
Answer:
[tex]P=6.10atm[/tex]
Explanation:
Hello,
In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:
[tex]PV=nRT[/tex]
Thus, since we are asked to compute the pressure y simply solve for it as follows:
[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]
Best regards.
9. Ibuprofen contain which of the following two functional groups: (1 point)
A) benzene
B) halogen
C) carboxyl
D) hydroxyl
Answer:
A and C
I hope this helps you:)
Choose the substance with the lowest boiling point.
A. NBr3.
B. CI2H2.
C. H2O2.
D. H2S.
E. O2.
Answer:
E. O2
Explanation:
All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.
In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.
Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.
Current is described as
A. moles of electrons.
B. the flow of electrons through a substance.
C. electricity.
D. the flow of ions through a substance.
Answer:
B!
Explanation:
I got it right in class!
interpret the electron configuration
Answer:
Ca for calcium
20 electrons
2-2s electron