Convert 6.21 km to m

Glucose, C6H12O6,C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the eq

Ilia_Sergeevich [38]

Answer:

Mass of oxygen = 61.824 g

Mass of carbon dioxide = 85.01 g

Explanation:

Given data:

Mass of glucose = 58 g

Mass of carbon dioxide = ?

Mass of oxygen = ?

Solution:

First of all we will write the balanced chemical equation,

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Moles of glucose:

Number of moles = mass / molar mass

Number of moles = 58 g/180 g/mol

Number of moles = 0.322 mol

Now we compare the moles of oxygen with glucose from balance chemical equation.

C₆H₁₂O₆ : O₂

1 : 6

0.322 : 0.322×6 = 1.932 mol

Mass of oxygen:

Mass of oxygen = number of moles × molar mass

Mass of oxygen = 1.932 mol × 32 g/mol

Mass of oxygen = 61.824 g

Now we compare the moles of carbon dioxide with moles of glucose and oxygen.

C₆H₁₂O₆ : CO₂

1 : 6

0.322 : 0.322×6 = 1.932 mol

O₂ : CO₂

6 : 6

1.932 : 1.932

Mass of carbon dioxide;

mass of carbon dioxide = number of moles × molar mass

mass of carbon dioxide = 1.932 mol × 44 g/mol

mass of carbon dioxide = 85.01 g

7 0

3 years ago

What mass in grams would 5.7L of hydrogen gas occupy at STP?

tekilochka [14]

Answer: The correct answer is: " 0.54 g " .

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Explanation:

Note that "hydrogen gas" is:

H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;

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The molecular weight of "H" is: 1.00794 g ;

(From the Periodic Table of Elements).

So, the molecular weight of: H₂ (g) is:

" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;

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Note the conversion for a gas at STP:

______

1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So: " 5.7 L H₂ (g) $* \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;$

The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

[5.7 * 2.104588 g ] / 22.4 = ? g ;

______________________

→ [ 11.9961516 g ] / 22.4 =

0.53554248214 g ;l

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We round this value to: " 0.54 g " ;

→ since "5.7 L " has 2 (two) significant figures;

22.4 is an exact number conversion;

and "5.7 L" has fewer significant figures than:

" 2.104588 " ; or: " 1.00794 " .

→ as such: We round to "2 (two) significant figures."

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Hope this is helpful. Wishing you the best in your academic endeavors!

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8 0

3 years ago

Nuclear energy comes from an atom's _____. <br><br> A. shells B. charge C. nucleus D. electrons

Shalnov [3]

This would be C. nucleus

4 0

3 years ago

Read 2 more answers

Help needed urgently<br>the word missing or partially seen is oxygen I forgot to put more pointa

butalik [34]

Explanation:

2.04 % hydrogen

32.65% sulphur

65.31% is oxygen

atomic ratio

hydrogen =2.04÷1=2.04

sulphur =32.65÷32=1.02

oxygen =65.31÷16=4.08

simplest ratio

hydrogen = 2.04÷1.02=2

sulphur =1.02÷1.02=1

oxygen =4.08÷1.02=4

empirical formula is H2SO4

4 0

3 years ago

Which atom would it be most difficult to remove an electron from

Norma-Jean [14]

I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon

6 0

3 years ago