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Tema [17]
3 years ago
13

in a reaction that occurs in a solution, the volume will not change. What happens to the concentration of the reactants? What ha

ppens to the reaction rate as the reaction continues?
Chemistry
1 answer:
mixer [17]3 years ago
8 0

Answer:

Concentration of the reactants decreases;

Rate of a reaction decreases

Explanation:

Since volume is held constant in a solution, the concentration of the reactants will decrease.

Simply speaking, concentration is the ratio between moles and volume. Volume is constant here. Over time, the number of moles of the reactants will decrease, as they react to produce products and they are disappearing. Since moles are directly proportional to concentration, this implies that concentration will also decrease, while the concentration of the products will increase, as they're formed.

The rate of a reaction decreases as the reaction continues, as we have lower and lower amounts of the reactants remaining in the solution as time progresses. Therefore, the probability of a successful collision leading to products decreases.

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
Hi mate <br><br> Pls help me the 8 one a <br><br> With lots of love: <br> Hareem
Sergeu [11.5K]

Answer:

Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the

3 0
3 years ago
Read 2 more answers
Please help if you know please and thanks ;)
GrogVix [38]

Answer:

it takes two hydrogen atoms for every oxygen atom to produce this reaction, so the mole ratio between hydrogen and oxygen is 2:1

8 0
3 years ago
The volume of a sample of water is 2.5 milliliters (mL). The volume of this sample in liters (L) is
Masteriza [31]

Answer:

2.5 Times . 10–3 L.

Explanation:

1 L = 1000 mL

1 mL = 1×10⁻³ L

2.5 mL / 1000 = 2.5×10⁻³ L

2.5 mL . 1×10⁻³ = 2.5×10⁻³ L

8 0
3 years ago
Read 2 more answers
100 points to the person who can answer these questions
olga55 [171]
I didn’t get to the the question, what was it?
3 0
3 years ago
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