Ask question

Ask question

All categories

3 years ago

8

Which sample of matter can be separated into different substances by physical means?

2 answers:

Art [367]3 years ago

6 0

Answer:

(1) LiCl(aq)

Explanation:

<u>Question:</u>

Which sample of matter can be separated into different substances by physical means?

1. LiCl(aq) ✔

2. LiCl(s) ❌

3. NH₃(g) ❌

4. NH₃(ℓ) ❌

<u>Reason:</u>

LiCl(a) is a mixture of a compound and water. The LiCl can be separated from the water. All the other choices are compounds.

<u>Related Vocabulary:</u>

- Aqueous: a water solution , indicated by the notion "(aq)". For example, HCl(aq) indicated a water solution of hydrochloric acid.

- Matter: Anything that has mass and takes up space.

Anna [14]3 years ago

4 0

3 since those are both solids that you could physical seperate. Liquids, gases, and aqueous materials have combined already

You might be interested in

61.0 mol of P4O10 contains how many moles of P

galina1969 [7]

61mol * 4 = 244moles of P

5 0

3 years ago

Read 2 more answers

dezoksy [38]

Answer:

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=

Answer: 9.04 g of H2O

Explanation:

First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)

Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)

Use equation to get moles and plug given

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O

3 0

2 years ago

A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid

const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

<u>Answer:</u> The pH of the solution is 10.09

<u>Explanation:</u>

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL$

Putting values in above equation, we get:

$1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL$

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For HClO:</u>

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

$0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol$

<u>For NaClO:</u>

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

$\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M$

To calculate the pH of the solution, we use the equation:

$pH=7+\frac{1}{2}[pK_a+\log C]$

where,

$pK_a$ = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

$pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09$

Hence, the pH of the solution is 10.09

4 0

3 years ago

How to use molar mass

ruslelena [56]

Explanation:

the molar mass of a compound can be caucaleted by adding the standar atomic masses.

8 0

3 years ago

Read 2 more answers

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is $1.743\times 10^2g$

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago