What is the mass of 4.02 mol Ba(NO2)2

Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h

andrew11 [14]

Answer : The change in entropy is $6.05\times 10^3J/K$

Explanation :

Formula used :

$\Delta S=\frac{m\times L_v}{T}$

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 1.00 kg

$L_v$ = heat of vaporization of water = $2256\times 10^3J/kg$

T = temperature = $100^oC=273+100=373K$

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

$\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}$

$\Delta S=6048.25J/K=6.05\times 10^3J/K$

Therefore, the change in entropy is $6.05\times 10^3J/K$

5 0

3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

2 years ago

1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by

Nitella [24]

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

6 0

3 years ago

3. What measurement unit is used to measure molecules

lakkis [162]

Answer: It is Mole.

Explanation:

Mole is the SI unit of measurement used to measure the number of things, usually atoms or molecules. One mole of something is equal to 6.02214078×1023 of same things (Avogadro's number).

8 0

3 years ago

Read 2 more answers

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

djyliett [7]

They are stronger than hydrogen bonding forces.

3 0

3 years ago

Read 2 more answers