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2 years ago

How many grams of glucose, C6H12O6, in 2.47 mole?

2 answers:

7 0

Answer:
number of grams = 444.6 grams

Explanation:
From the periodic table, we can find that:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
This means that:
molar mass of C6H12O6 = 6(12) + 12(1) + 6(16) = 180 grams

Now, number of moles can be calculated using the following rule:
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass = 2.47 * 180
mass = 444.6 grams

Hope this helps :)

statuscvo [17]2 years ago

6 0

First, we need to find the atomic mass of $C_{6}H_{12}O_{6}$ .

According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16

As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of   $C_{6}H_{12}O_{6}$ = 6 * 12.01 + 12 * 1.008 + 6 * 16

The molar mass of   $C_{6}H_{12}O_{6}$ = 180.156 grams/mole

Now that we have the molar mass of   $C_{6}H_{12}O_{6}$ , we can find the grams of glucose by using:

mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)

Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams

Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams

-i

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2 years ago

The chemical equation below shows the reaction of glucose and oxygen.

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2 years ago

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Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k = (2.303 /t) log [A₀] /[Aₙ]

where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, t

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substituting the values in the equation of the rate constant;

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2 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

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Answer:

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$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

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