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3 years ago

Elle wanted to test how temperature affects the rate of evaporation. She had three setups:

Chemistry

1 answer:

Pie3 years ago

5 0

Answer:

Temperature

Explanation:

Here the factor that Elle is controlling is the temperature. So temperature here is the independent variable and the dependent variable is the rate of evaporation of water. Independent variable is controlled during the experiment setup and the outcome of the dependent variable depends on the independent variable.

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3 years ago

What is the molar mass of Sr?

hichkok12 [17]

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3 years ago

How many grams of NO are produced from 15.8 g of H2O

Yuki888 [10]

Answer:

26.33 g of NO.

Explanation:

We'll begin by writing the balanced equation to produce NO from H2O. This is given below:

5N2 + 6H2O —> 4NH3 + 6NO

Next, we shall determine the mass of H2O that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 6 x 30 = 180 g

From the balanced equation above,

108 g of H2O reacted to produce 180 g of NO.

Finally, we shall determine the mass of NO produced by the reaction of 15.8 g of H2O.

This is illustrated below:

From the balanced equation above,

108 g of H2O reacted to produce 180 g of NO.

Therefore, 15.8 g of H2O will react to produce = (15.8 x 180)/108 = 26.33 g of NO.

Therefore, 26.33 g of NO were obtained from the reaction.

3 0

3 years ago

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$