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stepan [7]
3 years ago
7

A student wants to make a fair comparison of the reactivity of the metals with hydrochloric acid.

Chemistry
1 answer:
Alenkinab [10]3 years ago
4 0

Answer:

the concentration of the reactants

the temperature in heating

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Calculate the number of moles of BaF2 in a 10.0 g sample of BaF2.
UkoKoshka [18]

Answer:

The number of moles = 0.06 moles

Explanation:

i) Formula

Number of moles = Mass of the sample ÷ Molar masses of the substances

ii) Number of moles = 10.0 ÷ (137)+(19×2)

= 10.0 ÷ 175

= 0.06 moles

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3 years ago
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If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula
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Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.

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2 years ago
What is the mass of 7 x 10^28 atoms of Fe?
Nostrana [21]

Answer:

6 x 10⁶ g Fe

Explanation:

Step 1: Set up dimensional analysis

7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)

Step 2: Multiply, divide, and cancel out units

atoms Fe and atoms Fe cancel out.

mol Fe and mol Fe cancel out.

We should be left with g Fe.

7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe

116279(55.85) = 6.49 x 10⁶ g Fe

Step 3: Sig figs

There is only 1 sig fig in this problem.

6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe

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3 years ago
Describe heat transfer of Getting hot sitting outside in the sun.
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3 years ago
Consider the following chemical reaction: 2KCl + 3O2 --&gt; 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...
g100num [7]

Answer:

O₂; KCl; 33.3  

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

            2KCl  +  3O₂ ⟶ 2KClO₃

n/mol:  100.0   100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}

(ii) From O₂

\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0
3 years ago
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