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luda_lava [24]
2 years ago
8

Determine the number of moles of H in each sample

Chemistry
1 answer:
MissTica2 years ago
5 0

C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole

D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles

<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>

1 mole of N₂H₄ contains 4 moles of H

Therefore,

0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H

<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>

1 mole of C₁₀H₂₂ contains 22 moles of H

Therefore,

34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H

Learn more about mole:

brainly.com/question/13314627

#SPJ1

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The freezing point of pure water at sea level is 0°C. What is that in kelvins?
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Answer:b

Explanation:

The Kelvin degree is the same size as the Celsius degree; hence the two reference temperatures for Celsius, the freezing point of water (0°C), and the boiling point of water (100°C), correspond to 273.15°K and 373.15°K, respectively.

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Potassium and fluorine are both halogens?​
andreev551 [17]

Answer:

false, Potassium and fluorine are not halogens.

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What is the major difference, in terms of weathering type, between Albany, NY and Washington, D.C.? What might explain this diff
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Base your answers to the following questions on the graph below, which shows the effect that average yearly precipitation and temperature have on the type of weathering that will occur in a particular region.

City
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4 0
3 years ago
If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?
svet-max [94.6K]

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }

Where;

N(t) = The mass left after t years

N₀ =  The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

t_{\frac{1}{2} } = Half-life = 12.32 years

N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563

The mass left after 24.6 years = 25.0563 grams.

4 0
3 years ago
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