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Nat2105 [25]
3 years ago
12

If you have 5.22*1022 molecules of glucose (C6H12O6) how many grams is this?

Chemistry
1 answer:
lianna [129]3 years ago
7 0

Answer:

5.22*1022 molecules of glucose (C6H12O6)  is equal to 15.08 grams

Explanation:

Number of molecule of glucose in one mole = 6.023 * 10^{23}

Now mass of one mole of glucose = 180 grams

Number of moles in 6.023 * 10^{23} molecules of glucose = 1

Number of moles in 1 molecules of glucose =  \frac{1}{6.023 * 10^{23}}

Number of moles in 5.22 * 10^{22} molecules of glucose =  \frac{5.22* 10^{22}}{6.023 * 10^{23}}  = 0.083 moles

Weight of 0.083 moles

= 0.083 * 180\\= 15.08grams

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Answer:that sugar dissolves faster in a warm liquid than in a cool

Explanation:

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Changing the number of _____ would change an atom into an atom of a different element. changing the number of _____ would change
AURORKA [14]
Answer: Protons

WHY?
Changing electrons will only result in the same element having different charges and hence, changing it's chemical properties.

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3 years ago
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Read 2 more answers
g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole
Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

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3 years ago
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