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Rainbow [258]
2 years ago
13

when 1 mol of glucose is burned, 2802.5kj of energy is released. calculate rhe quantity of energy released to a person by eating

5.00g of glucose in a candy
Chemistry
1 answer:
tester [92]2 years ago
8 0

77.78 kJ of energy is released when 1 mol of glucose is burned, 2802.5 kJ of energy is released.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 ×10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Calculate the moles of 5.00g of glucose.

Given mass = 5.00g

The molar mass of glucose = 180.156 g/mol

Moles = \frac{mass}{molar \;mass}

Moles = \frac{5.00g}{180.156 g/mol}

Moles =0.02775372455

The quantity of energy released to a person by eating 5.00g of glucose in a candy.

0.02775372455 x 2802.5 kJ

77.77981305 kJ =77.78 kJ

Hence, 777.78 kJ of energy is released.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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Answer:

1.16 moles CO₂

Explanation:

To find the moles of CO₂, you need to (1) convert grams C₈H₁₈ to moles (via the molar mass) and then (2) convert moles C₈H₁₈ to moles CO₂ (via the mole-to-mole ratio from equation coefficients). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.

Molar Mass (C₈H₁₈): 8(12.011 g/mol) + 18(1.008 g/mol)

Molar Mass (C₈H₁₈): 114.232 g/mol

2 C₈H₁₈ + 25 O₂ -----> 16 CO₂ + 18 H₂O
^                                    ^

16.6 g C₈H₁₈            1 mole            16 moles CO₂          
--------------------  x  -----------------  x  -------------------------  = 1.16 moles CO₂  
                              114.232 g          2 moles C₈H₁₈          

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What amount of heat is required to raise the temperature of 20 grams of water from 10°C to 30°C? The specific heat of water is 4
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3 years ago
Read 2 more answers
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
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