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Galina-37 [17]
3 years ago
8

A chemist measures the energy change ?H during the following reaction:

Chemistry
1 answer:
Leno4ka [110]3 years ago
5 0

Answer:

(A) endothermic

(A) Yes, absorbed

Explanation:

Let's consider the following thermochemical equation.

2 Fe₂O₃(s) ⇒ 4 FeO(s) + O₂(g)  ΔH = 560 kJ

Since ΔH > 0, the reaction is endothermic.

We can establish the following relations:

  • 560 kJ are absorbed when 2 moles of Fe₂O₃ react.
  • The molar mass of Fe₂O₃ is 160 g/mol.

Suppose 66.6 g of Fe₂O₃ react. The heat absorbed is:

66.6g.\frac{1mol}{160g} .\frac{560kJ}{2mol} =117kJ

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How many moles are in 25.0 grams of KMnO4?
Lilit [14]

Answer:- 0.158 moles

Solution:- Moles are calculated on dividing the grams by the molar mass.

25.0 grams of KMnO_4 are given and we are asked to calculate the moles.

For this we need the molar mass so let's calculate it first.

Molar mass is the sum of atomic masses of all the atoms present in the molecule.

Molar mass = 39.098+54.938+4(16)

= 158.036 gram per mol

Now, we will divide the given grams by molar mass to get the moles and the set is shown as:

25.0gKMnO_4(\frac{1mol}{158.036g})

= 0.158molKMnO_4

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3 0
3 years ago
g What is the half-life for a particular reaction if the rate law is rate = (1291 M⁻¹*min⁻¹)[A]² and the initial concentration o
Gnoma [55]

Answer:

t_{1/2}=3.10x10^{-3}min=0.186s

Explanation:

Hello,

In this case, for the calculation of the half-life for the mentioned reaction we first must realize that considering the units of the rate constant and the form of the rate law, it is a second-order reaction, therefore, the half-life expression is:

t_{1/2}=\frac{1}{k[A]_0}

Therefore, we obtain:

t_{1/2}=\frac{1}{1291\frac{1}{M*min}*0.250M}\\\\t_{1/2}=3.10x10^{-3}min=0.186s

Regards.

4 0
4 years ago
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