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sdas [7]
2 years ago
10

The tabulated data show the rate constant of a reaction measured at several different temperatures. Use an Arrhenius plot to det

ermine the activation barrier and frequency factor for the reaction.
Temperature (K) Rate Constant (1/s)
300 0.0134
310 0.0407
320 0.114
330 0.303
340 0.757
​
Required:
Use an Arrhenius plot to determine the activation barrier for the reaction.
Chemistry
1 answer:
maxonik [38]2 years ago
5 0

Answer:

e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature

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Which of these is the largest quantity?
RideAnS [48]

Answer:

I would have to say the United States debt.

8 0
3 years ago
A sample of n2 effuses in 255 s. how long will the same size sample of cl2 take to effuse?
7nadin3 [17]
For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:

R₁/R₂ = √(M₂/M₁)

Let 1 be N₂ while 2 be Cl₂

255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s

<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
4 0
2 years ago
An independent variable is....
Irina18 [472]

Answer:

B - What we change

Explanation:

Dependent Variable - What we measure

Control Variable - what stays the same

Conclusion - what we conclude

<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>

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5 0
2 years ago
1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0
3 years ago
Question 2: Heat (5 points)
Ronch [10]

Answer:

(i) specific heat

(ii) latent heat of vaporization

(iii) latent heat of fusion

Explanation:

i. Q = mcΔT; identify c.

Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.  

ii. Q = mLvapor; identify Lvapor

Here, Q is the heat, m is the mass and L is the latent heat of vaporization.

The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.  

iii. Q = mLfusion; identify Lfusion

Here, Q is the heat, m is the mass and L is the latent heat of fusion.  

Here, Q is the heat, m is the mass and L is the latent heat of vaporization.

The amount of heat required to convert the 1 kg solid  into 1 kg liquid at constant temperature.  

8 0
3 years ago
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