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oee [108]
3 years ago
11

At a certain temperature this reaction follows first-order kinetics with a rate constant of 2.01 s^−1

Chemistry
1 answer:
ddd [48]3 years ago
6 0

Answer:

0.80 seconds (2 significant figures)

Explanation:

The equation of the reaction is given as;

CICH2CH2Cl (g) --> CH2CHCI (g) + HCl(g)

Rate constant (k) = 2.01 s^-1

From the units of the rate constant, this is a first order reaction.

Initial Concentration = 1.34 M

t = ?

Final concentration = 20% of 1.34 = 0.268 M

The integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln(0.268) = ln(1.34) - 2.01(t)

-2.01(t) = - 1.6094

t = 0.8007 ≈ 0.80 seconds (2 significant figures)

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Which of the following is the correct measurement for the volume of liquid shown below?
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B, 51 mL of mystery solution
4 0
3 years ago
How many copper atoms are in 70g of copper
Elanso [62]

Answer:

n = 6.634\times 10^{23}\,atoms

Explanation:

The total amount of atoms is found by multiplying Avogradro's Number and the number of moles:

n = \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)\cdot \left(\frac{70\,g}{63.546\,\frac{g}{mol} } \right)

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4 0
4 years ago
How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0
3 years ago
Jacob is doing a report on the galaxy in which Earth is found. Which of the following is the galaxy that Jacob should research f
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Answer:

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3 0
3 years ago
Acetylene gas (c2h2) reacts with oxygen gas (o2) to produce carbon dioxide (co2) and water vapor (h2o). how many liters of c2h2
MissTica
According to the balanced equation of the reaction:

2C2H2 + 5O2 → 4CO2 + 2H2O

So we can mention all as liters,

A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O 

So, when we have 75L of CO2

and when we have 2 L of C2H2 reacts and gives 4 L of CO2

 2C2H2 → 4CO2 

∴ The volume of C2H2 required is:

                    = 75L / 2

                    = 37.5 L

B) and, when we have 75 L of CO2


and 4CO2 → 2H2O 

∴ the volume of H2O required is: 
                               
                            = 75 L /2
                            = 37.5 L 


C) and from the balanced equation and by  the same way:

when 5 liters O2  reacts to give 4 liters of CO2

and we have 75 L of CO2:

5 O2 → 4 CO2 

   ??  ←  75 L

∴ the volume of O2 required is:

                              = 75 *(5/4)

                              = 93.75 L



D) about the using of the number of moles the answer is: 

no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
6 0
3 years ago
Read 2 more answers
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