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3 years ago

How many moles of h are in 0.73 mole of c6h10s

Chemistry

1 answer:

Mnenie [13.5K]3 years ago

7 0

Answer:

7.3 mole

Explanation:

1 C6H10S gives 10H

0.73 mole of C6H10S gives X mole of Hydrogen

X= 0.73×10=7.3mole

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Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence. The early atmosphere was probably mostly carbon dioxide, with little or no oxygen.

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3 years ago

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3 years ago

Lokesh is sawing boards to make a deck. What is the area of the deck?

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Explanation:

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3 years ago

Read 2 more answers

when nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic

AnnZ [28]

Using stoichiometry, 3.24 grams $NaHCO_{3}$ will yield 2.04 grams $Na_{2} CO_{3}$ and 1.20 grams $H_{2} CO_{3}$ . But in an actual decomposition where the mass of one of the products was measured to be 2.01 grams, which is $Na_{2} CO_{3}$ , the percent yield is 98.34%.

Stoichiometry is defined as the relationship between the amounts of substances that are involved in chemical reactions.

Mole ratio relates the amounts in moles of any two substances by looking at the coefficients in front of each species in the balanced chemical equation.

For the decomposition of $NaHCO_{3}$ , it follows the balanced chemical equation:

$2NaHCO_{3} = Na_{2} CO_{3}+H_{2} CO_{3}$

2 moles of $NaHCO_{3}$ will decompose into 1 mole of $Na_{2} CO_{3}$ and 1 mole of $H_{2} CO_{3}$

mole ratio = 2 : 1 : 1

If the mass of $NaHCO_{3}$ sample is 3.24 grams, then

molar mass = m/n

84.007 g/mol = 3.24 g/n

n = 0.03856821455 mol $NaHCO_{3}$

and using the mole ratio 2 : 1 : 1, theoretically,

0.03856821455 mol $NaHCO_{3}$ will yield 0.01928410728 mol $Na_{2} CO_{3}$ and 0.01928410728 mol $H_{2} CO_{3}$ .

If there's 0.01928410728 mol $Na_{2} CO_{3}$ , its mass is:

molar mass = m/n

105.9888 g/mol = m/ 0.01928410728

m = 2.043899389 g $Na_{2} CO_{3}$

On the other hand, if there's 0.01928410728 mol $H_{2} CO_{3}$ , then

molar mass = m/n

62.03 g/mol = m/0.01928410728

m = 1.196193174 g $H_{2} CO_{3}$

If the mass of one of the products was measured to be 2.01 grams, then it must be $Na_{2} CO_{3}$ since the experimental value is much closer to the theoretical mass.

Calculating its percent yield,

percent yield = actual yield/theoretical yield x 100%

percent yield = 2.01 g/2.043899389 g x 100%

percent yield = 98.34143552% = 98.34%

To learn more about stoichiometry: brainly.com/question/25829169

#SPJ4

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2 years ago

What mass, in grams, of sodium bicarbonate, nahco3, is required to neutralize 1000.0 l of 0.350 m h2so4?

oksano4ka [1.4K]

Hello!

The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:

2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂

To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:

$1000 L H_2SO_4* \frac{0,350 moles H_2SO_4}{1 L}* \frac{2 moles NaHCO_3}{1 mol H_2SO_4}* \frac{84,007gNaHCO_3}{1molNaHCO_3} \\ \\ =58804,9 g NaHCO_3$

So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃

3 0

3 years ago