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vladimir2022 [97]
3 years ago
9

1. Which of the following choices demonstrates the law of constant composition? (Slides 2 ‒ 3: Laws on Matter) (a) Nitrogen and

oxygen are both found in nature as diatomic molecules. (b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen. (c) Nitrogen and oxygen gases are mixed to produce a sample consisting of 20.0 g of nitrogen and 32.0 g of oxygen. (d) When 14.0 g of nitrogen and 32.0 g of oxygen are mixed and react, there is no measurable change in mass during the reaction. (e) Nitrogen and oxygen are gases found in air, nitrogen is approximately 79% of air and oxygen is approximately 21%.
Chemistry
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

(b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen.

Explanation:

The law of definite proportion states that a gen chemical compound always contains its constituent elements in a fixed ratio by mass, independent on the method of preparation.

The molar mass of Nitrogen and Oxygen would always remain the same, allowing for exact reactant masses (or mole ratio) irrespective of the given amount of sample.

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Leokris [45]
Answer: 1.03 x 10^25 ions Na
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3 years ago
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Never been taught this. Need help.
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Definitely not D i would guess A or B
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g The reaction; 4 Ag(s) + O2(g) ----> 2 Ag2O(s), is exothermic. Which statement about the reaction is correct? (A) It is spon
melomori [17]

Answer:

The correct answer is B. It is spontaneous only at low temperatures.

Explanation:

In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.

The spontaneity of a reaction is given by the equation:

ΔG = ΔH - TΔS

where:

ΔH: enthalpy variation

T: absolute temperature

ΔS: entropy variation

As the reaction is exothermic, ΔH<0

As the reaction order increases (the reagents are solid and gas and their product is solid), ΔS<0

Therefore, the reaction will be spontaneous when ΔG is negative.

ΔG = ΔH - TΔS

That is, the entropy term must be smaller than the enthalpy term.

Hence, the reaction will be spontaneous only at low temperatures.

4 0
3 years ago
Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made
arlik [135]

Answer: The empirical formula is C_3H_3O.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2 = 12.24 g

Mass of H_2O = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

5 0
3 years ago
A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of
mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.125

t = ?

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.125=e^{-0.00440\times t}

t = 473 year

8 0
3 years ago
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