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ArbitrLikvidat [17]
2 years ago
9

Which of the following is an example of a phase change from liquid to gas?

Chemistry
2 answers:
sesenic [268]2 years ago
8 0
C.Water evaporating is the correct answer
You can go for it
netineya [11]2 years ago
6 0
Correct answer to this question is c. water evaporating
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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent
elixir [45]

Answer : The initial rate for a reaction will be 3.8\times 10^{-4}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

A+B+C\rightarrow P

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1

k=7.6\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1

\text{Rate}=3.8\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.8\times 10^{-3}Ms^{-1}

6 0
3 years ago
Which of the following is NOT a reason why chemical bonds are important? *
KengaRu [80]

Answer:

the answer would be B :)

3 0
3 years ago
4. What is the volume of a box that has a length of 50cm, a width of 20cm,<br> and a height of 5 cm?
EastWind [94]

5,000 cm3 (cubic centimeters)

4 0
2 years ago
What contains most of earths water
nikitadnepr [17]
The ocean hold 97 percent of earths water.
5 0
3 years ago
0.1005 liters is the same as: A. 0.0001005 cm3 B.0.1005 cm3 C.100.5 cm3 D.0.01005 cm3 and A. 0.01005 mL B. 0.1005 mL C. 0.000100
Andreyy89

I think it would be C.100.5cm or D.100.5ml hope that helps


4 0
3 years ago
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