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3 years ago

How many atoms are in 0.23 moles of calcium?

Chemistry

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

1.385 × 10²³ atoms of calcium

Explanation:

Given data:

Number of moles of calcium = 0.23 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 0.23 mole of calcium:

one mole = 6.022 × 10²³ atoms of calcium

0.23 × 6.022 × 10²³ atoms of calcium

1.385 × 10²³ atoms of calcium

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Answer:

The correct answer is option D.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

$NO_2(g) + CO(g)\rightarrow NO(g) + CO_2(g)$

Rate of the reaction:

$R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}$

Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.

Given rate expression of the reaction:

$R = k[NO2]^2[CO]$

Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'

$R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R$

Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.

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2 years ago

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A sample of a chromium-containing alloy weighing 3.450 g was dissolved in acid, and all the chromium in the sample was oxidized

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Answer:

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Explanation:

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Oxidation half reaction

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2 years ago

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2 years ago

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Ksenya-84 [330]

Answer
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Explanation

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Consider the equation:

Dmitry [639]

Answer:1. $Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. The rate constant (k) for the reaction is $3.50M^\frac{-1}{2}s^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$rate=k[CHCl_3]^x[Cl_2]^y$

k= rate constant

x = order with respect to $CHCl_3$

y = order with respect to $Cl_2$

n = x+y= Total order

1. a) From trial 1: $0.0035=k[0.010]^x[0.010]^y$ (1)

From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (2)

Dividing 2 by 1 : $\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}$

$2=2^x,2^1=2^x$ therefore x=1.

b) From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (3)

From trial 3: $0.0098=k[0.020]^x[0.020]^y$ (4)

Dividing 4 by 3: $\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}$

$1.4=2^y,2^{\frac{1}{2}}=2^y$ therefore $y=\frac{1}{2}$

$rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. to find rate constant using trial 1:

$0.0035=k[0.010]^1[0.010]^\frac{1}{2}$

$k=3.50M^\frac{-1}{2}s^{-1}$

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