All categories

3 years ago

Can someone help me with this?

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

4 0

Answer:

idk my man srry5678976543567890876546789087659687876875787678687367376367373

Explanation:

You might be interested in

What solid materials were dissolved in water?

Marat540 [252]

Explanation:

salt, sugar and coffee :)

5 0

2 years ago

Read 2 more answers

A gas occupies a volume of 160 ml at 36° C and 970mm Hg. What is the volume at STP?

igor_vitrenko [27]

Answer:

I'm sorry I just need points

Explanation:

7 0

2 years ago

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re

lord [1]

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

7 0

3 years ago

Read 2 more answers

A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will

Reil [10]

Answer:

The balanced chemical equation for this process:

$H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)$

$194.51 is the cost that the firm will incur from this use of slaked lime.

Explanation:

The balanced chemical equation for this process

$H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)$

Moles of sulfuric acid = n

Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L

1 gallon = 3.785 Liter

Morality of the sulfuric acid = M = 1.05 m

$Molarity(M)=\frac{n}{V(L)}$

$n=m\times V=1.05 M\times 4,542 L=4,769.1 mol$

According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;

$\frac{1}{1}\times 4,769.1 mol=4,769.1 mol$ of calcium hydroxide

Mass of 4,769.1 moles of calcium hydroxide:

4,769.1 mol 74 g/mol = 352,913.4 g

= $\frac{352,913.4 }{453.6} pounds =778.03 pounds$

(1 pound = 453.6 grams)

Cost of 1 pound of slaked lime = $0.25

Cost of 778.03 pounds of slaked lime = $0.25 × 778.03 = $194.51

$194.51 is the cost that the firm will incur from this use of slaked lime.

3 0

3 years ago

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$ .

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of $NiSO_4$ .

Molar mass of $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$ :

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$ :

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$ :

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$

3 0

3 years ago