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MrMuchimi
3 years ago
12

A 3.60−g sample of a salt dissolves in 7.70 g of water to give a saturated solution at 21°C. What is the solubility (in g salt/1

00 g of H2O) of the salt?
Chemistry
1 answer:
laiz [17]3 years ago
7 0

Answer:

The solubility of the salt is

46.75g salt/100g of H2O

Explanation:

If 7.70g of water can dissolve 3.60g at 21°C

That is 1.0g of water can dissolve (3.6÷7.7) grams of salt, that is 0.4675_g of salt at 21°C or 100 grams of water can dissolve 100×0.4675 = 46.75g

The solubility of the salt is

46.75g salt/100g of H2O

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​What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water? 
dusya [7]

Answer:

The concentration of KBr is  C = 0.07899 \ mol   L^{-1}

Explanation:

From the question we are told that

      The mass of KBr is  m_{KBr}  = 2.35 \ g

       The molar mass of KBr is  M_{KBr} =  119 g/mol

       Volume of water is V = 250 \ mL = 250 *10^{-3} =  0.250 \ L

This implies that the volume of  the solution is  V = 250 mL

The number of moles of KBr is

         n = \frac{m_{KBr}}{M_{KBr}}

Substituting values

         n =  \frac{2.35}{119}

        n = 0.01975 \ mol

The concentration of KBr is mathematically represented as

                C = \frac{0.01975}{0.250}

                C = 0.07899 \ mol   L^{-1}

3 0
3 years ago
Which increases the rate of a chemical reaction?
Anastasy [175]
A catalyst

A catalyst can be in many forms
3 0
3 years ago
A student sets up the following equation to convert a measurement.
dybincka [34]

Answer:

\frac{1 m}{100 cm}

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

<h3>-1.7*10^5\frac{V}{m}  ·  \frac{1 m}{100 cm}  =  ? \frac{V}{cm}</h3><h2>                 ^</h2>

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0
2 years ago
A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure o
larisa86 [58]

Answer:

  • <em>The partial pressure of oxygen in the mixture is</em><u> 320.0 mm Hg</u>

Explanation:

<u>1) Take a base of 100 liters of mixture</u>:

  • N: 60% × 100 liter  = 60 liter

  • O: 40 % × 100 liter = 40 liter.

<u>2) Volume fraction:</u>

At constant pressure and temperature, the volume of a gas is proportional to the number of molecules.

Then, the mole ratio is equal to the volume ratio. Callin n₁ and n₂, the number of moles of nitrogen and oxygen, respectively, and V₁, V₂ the volume of the respective gases you can set the proportion:

  • V₁ / V₂ = n₁ / n₂

That means that the mole ratio is equal to the volume ratio, and the mole fraction is equal to the volume fraction.

Then, since the law of partial pressures of gases states that the partial pressure of each gas is equal to the mole fraction of the gas multiplied by the total pressure, you can draw the conclusion that the partial pressure of each gas is equal to the volume fraction of the gas in the mixture multiplied by the total pressure.

Then calculate the volume fractions:

  • Volume fraction of a gas = volume of the gas / volume of the mixture

  • N: 60 liter / 100 liter = 0.6 liter

  • V: 40 liter / 100 liter = 0.4 liter

<u>3) Partial pressures:</u>

These are the final calculations and results:

  • Partial pressure = volume fraction × total pressure

  • Partial pressure of N = 0.6 × 800.0 mm Hg = 480.0 mm Hg

  • Partial pressure of O = 0.4 × 800.0 mm Hg = 320.0 mm Hg
8 0
3 years ago
Using the balanced equation below, how many moles of water can be produced when 4 grams of Hydrogen react? 2H2(g)+O2(g) —&gt; 2H
gavmur [86]

Answer:

4 × 10 g

Explanation:

Step 1: Write the balanced equation

2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)

Step 2: Calculate the moles corresponding to 4 g of H₂

The molar mass of H₂ is 2.02 g/mol.

4 g × 1 mol/2.02 g = 2 mol

Step 3: Calculate the moles of H₂O produced from 2 moles of H₂

The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.

Step 4: Calculate the mass corresponding to 2 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

2 mol × 18.02 g/mol = 4 × 10 g

7 0
3 years ago
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