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Kitty [74]
3 years ago
13

Please please help me solve this, I have a test and only 20 mins left T-T​

Chemistry
1 answer:
Ann [662]3 years ago
7 0

Answer: 5.6dm^3[/tex[ of gas is produced when 0.1 moles of magnesium nitrate is decomposed.Explanation:The balanced chemical equation is:
[tex]2NH_4NO_3(s)\rightarrow 2MgO(s)+4NO_2(g)+O_2(g)

According to stoichiometry :

2 moles of NH_4NO_3 produce = 4 moles of NO_2 gas and 1 mole of O_2 gas

2 moles of NH_4NO_3 produce = 5 moles of gas

Thus 0.1 mole of NH_4NO_3 produce =\frac{5}{2}\times 0.1=0.25moles  of gas

Volume of gas produced = moles\times {\text {Molar volume}}=0.25moles\times 22.4dm^3/mol=5.6dm^3

Thus [tex]5.6dm^3[/tex[ of gas is produced when 0.1 moles of magnesium nitrate is decomposed.

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Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−
Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 9.7\times 10^{-6}M^{-1}s^{-1}

t = time taken  = 5.00 days

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[A]_o = Initial concentration = 0.110 M

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

9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)

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8 0
3 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

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

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
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