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3 years ago

How many molecules are there in 79g of Fe2O3? how many atoms is thi

1 answer:

Nookie1986 [14]3 years ago

7 0

Molar mass of iron(III) oxide is 160g/mol

160g ----------- of iron(III) oxide is ---------------- 6.02*10²³molecules
79g ----------- of iron(III) oxide is ---------------- x

x = [6.02*10²³*79g]/160g ≈ <u>2.97*10²³ molecules</u>

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An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not

Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

$5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} )$ = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0

2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

2 years ago

Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:

Lubov Fominskaja [6]

Answer:

$\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}$

Explanation:

(a) Balanced equation

Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag

(b) Calculation

You want to convert atoms of Cu to atoms of Ag.

The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

$\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}$

4 0

2 years ago

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Yakvenalex [24]

Answer:

mitochondrion

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3 years ago

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Kamila [148]

The idea of electric field was presented by Michael Faraday. The electrical field constrain acts between two charges, similarly that the gravitational field compel acts between two masses.

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3 years ago