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frosja888 [35]
2 years ago
9

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

ation.) How many atoms will there be in 5.00 g of gold.
Chemistry
1 answer:
Kobotan [32]2 years ago
5 0

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

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Answer : The value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

Explanation :

The following equilibrium reactions are :

(1) 2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2 K_1=5.40\times 10^{-16}

(2) 2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l) K_2=1.06\times 10^{10}

(3) CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g) K_3=2.68\times 10^{-9}

The final equilibrium reaction is :

CO_2(g)\rightleftharpoons C(s)+O_2(g) K_{goal}=?

Now we have to calculate the value of K_{goal} for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

K_{goal}=\sqrt{K_1\times K_2\times K_3}

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

K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}

K_{goal}=1.238\times 10^{-7}

Therefore, the value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

3 0
3 years ago
A chemistry student weighs out 0.0475g of sulfurous acid H2SO3, a diprotic acid, into a
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MgO + H2O = Mg(OH)2 is an example of a _____ chemical reaction.
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a) synthesis- or combination- reaction

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2 years ago
What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?​
Semmy [17]

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

8 0
3 years ago
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