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3 years ago

1) 5,730 s to ks

Chemistry

2 answers:

Ksju [112]3 years ago

5 0

Answer: cubic metres (m3)

metres (m)

Hope this helped!!!

Len [333]3 years ago

4 0

Explanation:

1000 seconds =1ks

5730 seconds=?

5730×1/1000

5.73ks

10mg=1cg

22000mg?

22000×1/10

2200cg

1000ms=1ks

0.077ms=1ks

0.077×1/1000

0.000077ks

cubic metres (m3)

metres (m)

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When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

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Answer:

7.5 mol of hydrogen would be needed to consume the available nitrogen.

Explanation:

The balanced equation of the reaction of hydrogen with nitrogen is:

3H₂ + N₂ → 2NH₃

it is clear that 3mol of hydrogen react with 1 mol of Nitrogen to give 2 mol of ammonia.

firstly, we need to calculate the number of moles of both

for H₂:

number of moles = mass / molar mass = (7.00 g) / (2.00 g/mol) = 3.5 mol.

for N₂:

number of moles = mass / molar mass = (70.00 g) / (28.00 g/mol) = 2.5 mol.

using cross multiplication

1 mol of N₂ needs → 3 mol of H₂

2.5 mol of N₂ needs → ??? mol of H₂

∴ the number of mol of H₂ needed = (3*2.5) / 1 = 7.5 mol

So, the right choice is:

7.5 mol of hydrogen would be needed to consume the available nitrogen.

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KonstantinChe [14]

Answer: The molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L

Explanation:

We are given:

Solubility of nitrogen gas in water = 56.0 mg/100 g

Or, solubility of nitrogen gas in water = 0.056 g/100 mL (Density of water = 1 g/mL & Conversion factor used: 1 g = 1000 mg)

Solubility of a solute is defined as the moles of solute dissolved in 1 L of solvent.

Conversion factor used: 1 L = 1000 mL

Applying unitary method:

In 100 mL water, the amount of solute (nitrogen gas) dissolved is 0.056 grams

So, in 1000 mL of water, the amount of solute (nitrogen gas) dissolved will be = $\frac{0.056}{100}\times 1000=0.56g$

Converting this solubility into mol/L by dividing with the molar mass of nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

So, Solubility of nitrogen gas = $\frac{0.56g/L}{28g/mol}=0.2mol/L$

To calculate the Henry's constant we use the equation given by Henry's law, which is:

$C_{N_2}=K_H\times p_{N_2}$ .........(1)

where,

$K_H$ = Henry's constant

$C_{N_2}$ = molar solubility of nitrogen gas = 0.02 mol/L

$p_{N_2}$ = partial pressure of nitrogen gas = 2.38 atm

Putting values in equation 1, we get:

$0.02mol/L=K_H\times 2.38atm\\\\K_H=\frac{0.02mol/L}{2.38atm}=8.40\times 10^{-3}mol/L.atm$

When pressure is changed to 5.00 atm

Now,

$p_{N_2}=5.00atm\\\\K_H=8.40\times 10^{-3}mol/L.atm$

Putting values in equation 1, we get:

$C_{N_2}=8.40\times 10^{-3}mol/L.atm\times 5.00atm=0.042mol/L$

Hence, the molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L

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Answer:

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Explanation:

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