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3 years ago

Which compound of gold will produce the least mass of gold?

Chemistry

1 answer:

miss Akunina [59]3 years ago

7 0

Au₂S₃

Explanation:

Given compounds:

Gold(I) oxide = Au₂O

Gold(I)sulfide = Au₂S

Gold(III) oxide = Au₂O₃

Gold(III)sulfide = Au₂S₃

Problem: compound with the least mass of Gold.

To solve this problem, we simply find the mass percentage of gold in the given compounds.

Molar mass of Au₂O = 2(197) + 16 = 410g/mole

Molar mass of Au₂S = 2(197) + 32 = 426g/mole

Molar mass of Au₂O₃ = 2(197) + 3(16) = 442g/mole

Molar mass of Au₂S₃ = 2(197) + 3(32) = 490g/mole

The molar mass of gold in all the compound is 394g/mole

Mass percentage of gold in Au₂O = $\frac{394}{410}$ x 100 = 96.1%

Mass percentage of gold in Au₂S = $\frac{394}{426}$ x 100 = 92.5%

Mass percentage of gold in Au₂O₃ = $\frac{394}{442}$ x 100 = 89.1%

Mass percentage of gold in Au₂S₃ = $\frac{394}{490}$ x 100 = 80.4%

Au₂S₃ is the least mass of gold in it

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In a solution, what term is given to the liquid? a.Solute b.Solvent c.Reagent

PSYCHO15rus [73]

Answer:

B solvent

Explanation:

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3 years ago

Technetium-104 has a half-life of 18.0 minutes. How much of a 165 g sample is left after 90.0 minutes?

asambeis [7]

Answer:

After 90 minutes 5.15625 g amount left from total of 165 g.

Explanation:

Given data:

Technetium-104 half life = 18.0 min

Total amount of sample = 165 g

Amount left after 90.0 min = ?

Solution:

First of all we will calculate the number of half lives passes.

Number of half lives = T elapsed / Half life

Number of half lives = 90 min /18.0 min

Number of half lives = 5

Amount left:

At time 0 = 165 g

At first half life = 165 g/ 2= 82.5 g

At 2nd half life = 82.5 g/2 = 41.2 g

At 3rd half life = 41.2 g/ 2 = 20.625 g

At 4th half life = 20.625 g/ 2 = 10.3125 g

At 5th half life = 10.3125 g /2 = 5.15625 g

Thus, after 90 minutes 5.15625 g amount left from total of 165 g.

5 0

3 years ago

Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of

Natalija [7]

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = $\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}$

First we calculate the initial concentration:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We put the data in the Kc expression and solve for x:

$\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}$

$\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}$

We make a simplification because x<<< 0.0225:

$\frac{4x^3}{0.0225} =9.30*10^{-8}$

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

5 0

3 years ago

Read 2 more answers

I don't understand this I'm confused

skad [1K]

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4 years ago

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Setler79 [48]

Answer:

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3 0

3 years ago