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3 years ago

What effect would a decrease or increase in barometric pressure have on the boiling point

Chemistry

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

Pressure Affects the Boiling Point

Atmospheric pressure influences the boiling point of water. When atmospheric pressure increases, the boiling point becomes higher, and when atmospheric pressure decreases (as it does when elevation increases), the boiling point becomes lower.

Explanation:

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Commercial concentrated aqueous ammonia is 28% nh3 by mass and has a density of 0.90 g/ml. what is the concentration of ammonia

patriot [66]

Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

$100g(\frac{1mL}{0.90g})$

= 111 mL

Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

$111mL(\frac{1L}{1000mL})$

= 0.111 L

Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

$28g(\frac{1mole}{17g})$

= 1.65 mole

To calculate the molarity we divide the moles of ammonia by the liters of solution:

$molarity=\frac{1.65mole}{0.111L}$

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

5 0

4 years ago

An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

3 years ago

A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate

Paul [167]

Answer:

0.08 mol L-1

Explanation:

Sulfuric acid Formula: H2SO4

Ammonia Formula: NH3

Ammonium sulfate Formula: (NH₄)₂SO₄

H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

7 0

3 years ago

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0

3 years ago

How many moles of PC15 can be produced from 58.0 g of Cl₂ (and excess<br> P4)?

ludmilkaskok [199]

0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess

P4)

<h3>How to calculate moles?</h3>

The balanced chemical equation is

$P_{4} + 10Cl_{2} = 4PCl_{5}$

The mass of clorine is m( $Cl_{2}$ ) = 58.0 g

The amount of clorine is n( $Cl_{2}$ ) = m( $Cl_{2}$ )/M( $Cl_{2}$ ) = 58/70.906 = 0.817 mol

The stoichiometric reaction,shows that

10 moles of $Cl_{2}$ yield 4 moles of $PCl_{5}$ ;

0.817 of $Cl_{2}$ yield x moles of $PCl_{5}$

n( $PCl_{5}$ ) = 4*0.817/10 = 0.3268 mol

To know more about stoichiometric reaction, refer:

brainly.com/question/14935523

#SPJ9

3 0

2 years ago