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3 years ago

How much energy is required to vaporize 0.5kg of water?

Chemistry

1 answer:

White raven [17]3 years ago

3 0

Answer:

Explanation:

Vaporization Equation => Q = m·ΔHv

m = 0.5Kg water = 500g water

ΔHv = Heat of vaporization = 540 calories/gram

∴Q = 500g x 540cal/g = 270,000 calories x 4.184 joules/cal = 1,129,680 joules = 1130 Kilojoules (3 sig. figs.)

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When combining acids and water, water should always be added to the acid. True or False?

scoundrel [369]

Answer:

False

Explanation:

Always add acid to the water, it dilutes the acid the other way around.

8 0

2 years ago

If you mixed three Dixie cups’ contents together containing 0.05L of 1.0M lemonade, 0.05L of 2.5M lemonade, and 0.05L of 0.5M le

viva [34]

Molarity of the resulting solution will be 1.33 M.

<u>Explanation:</u>

First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume

For cup 1 = 1 M ×0.05 L = 0.05 moles

For cup 2 = 2.5 M × 0.05 L= 0.125 moles

For cup 3 = 0.5 M × 0.05 L = 0.025 moles

Total moles = 0.05 + 0.125 + 0.025 = 0.2 moles

We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L

Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M

3 0

2 years ago

What is the molarity of a solution containing 9.0 moles of solute in 2500 ml of a solution

Lena [83]

M=(mols/L) so M=(9.0/2.5) which is 3.6M

7 0

2 years ago

Which of the following is an example of a colloid

Ivan

I think it is mayonnaise

4 0

2 years ago

Read 2 more answers

NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?

Anna007 [38]

Answer: 25.13 g of $N_2O$ and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles$

The balanced chemical equation is:

$NH_4NO_3\rightarrow N_2O+2H_2O$

According to stoichiometry :

1 mole of $NH_4NO_3$ produce = 1 mole of $N_2O$

Thus 0.571 moles of $NH_4NO_3$ will require= $\frac{1}{1}\times 0.571=0.571moles$ of $N_2O$

Mass of $N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g$

1 mole of $NH_4NO_3$ produce = 2 moles of $H_2O$

Thus 0.571 moles of $NH_4NO_3$ will require= $\frac{2}{1}\times 0.571=1.142moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g$

Thus 25.13 g of $N_2O$ and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$

5 0

2 years ago