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3 years ago

Convert 6.93 x 1024 atoms of carbon to moles of carbon.

Chemistry

1 answer:

Aleksandr [31]3 years ago

7 0

Answer: 11.5 moles of carbon

Explanation:

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms

So, 1 mole of carbon = 6.02 x 10^23 atoms

Z moles = 6.93 x 10^24 atoms

To get the value of Z, cross multiply:

(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)

6.93 x 10^24 = (6.02 x 10^23 x Z)

Z = (6.93 x 10^24) ➗ (6.02 x 10^23)

Z = 1.15 x 10

Z = 11.5 moles

Thus, there are 11.5 moles of carbon.

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La temperatura de congelación del agua es 0 grados centígrados.

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3 years ago

Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is

Korolek [52]

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

$\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2$

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

Method 1: using the mass of glucose

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= $\frac{10}{9}$ mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

= $2(\frac{10}{9} )$

= $\frac{20}{9}$ mol

Mass of ethanol

= $\frac{20}{9} \times[2(12)+6+16]$

= $\frac{20}{9}(46)$

= 102 g (3 s.f.)

Method 2: using mass of carbon dioxide produced

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= $\frac{977}{440}$ mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= $\frac{977}{440}$ mol

Mass of ethanol formed

= $\frac{977}{440} \times[2(12)+6+16]$

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

brainly.com/question/1685725

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1 year ago

The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

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Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

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Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

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$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

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