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2 years ago

What reaction is depicted by the given equation: Au3+ + 3e− Au

1 answer:

4 0

The reaction represents an oxidation-reduction reaction where it shows the reduction reaction part. The Au is reduced from a charge of positive 3+ to a neutral charge. An oxidation-reduction reaction is any reaction that involves changes in the oxidation number by gaining or losing electrons.

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A flexible container at an initial volume of 3.10 L contains 9.51 mol of gas. More gas is then added to the container until it r

Schach [20]

Answer:

49.09 moles of gas are added to the container

Explanation:

Step 1: Data given

Initial volume = 3.10 L

Number of moles gas = 9.51 moles

The final volume = 19.1 L

The pressure, temperature remain constant

Step 2: Calculate number of moles gas

V1/n1 = V2/n2

⇒with V1 = the initial volume of the gas = 3.10 L

⇒with n1 = the initial number of moles = 9.51 moles

⇒with V2 = the increased volume = 19.1 L

⇒with n2 = the final number of moles gas

3.10L / 9.51 moles = 19.1 L / n2

n2 = 58.6 moles

The new number of moles is 58.6

Step 3: calculate the number of moles gas added

Δn = 58.6 - 9.51 = 49.09 moles

49.09 moles of gas are added to the container

7 0

3 years ago

How many grams of sulfuric acid are needed to produce 57.0 grams of water? Show all steps of your calculation as well as the fin

antiseptic1488 [7]

Answer:

155.16 g.

Explanation:

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2NaOH + H₂SO₄ → 2Na₂SO₄ + 2H₂O

It is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.

We must convert the grams of water (57.0 g) to moles (n = mass/molar mass).

n = (57.0 g) / (18.0 g/mole) = 3.1666 moles.

Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.

Using cross multiplication:

1.0 mole of H₂SO₄ → 2.0 moles of H₂O, from the stichiometry of the balanced equation.

??? moles of H₂SO₄ → 3.1666 moles of H₂O.

The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O (57.0 g) is (1.0 x 3.1666 / 2.0) = 1.5833 moles.

Finally, we should convert the number of moles of H₂SO₄ into grams (n = mass/molar mass).

Molar mass of H₂SO₄ = 98.0 g/mole.

mass = n x molar mass = (1.5833 x 98.0) = 155.16 g.

6 0

3 years ago

Read 2 more answers

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide (C6H12O6 + 6 O2 →

KATRIN_1 [288]

Answer:

what are your statements?

3 0

2 years ago

Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi

erastovalidia [21]

Answer:

The answer to your question is:

Explanation:

Data

carbon 7.3% = 7.3g

hydrogen 4.5% = 4.5g

oxygen 36.4% = 36.4 g

nitrogen 31.8% = 31.8 g

Now

For carbon

12 g --------------------1 mol

7.3 g ------------- x

x = 7.3/12 = 0.608 mol

For hydrogen

1 g -------------------- 1 mol

4.5 g ------------------ x

x = 4.5 mol

For oxygen

16 g ------------------- 1 mol

36.4 g ---------------- x

x = 2.28 mol

For nitrogen

14 g ---------------- 1 mol

31.8 g --------------- x

x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon 0.608/0.608 = 1

hydrogen 4.5/ 0.608 = 7.4

oxygen 2.28/0.608 = 3.75

nitrogen 2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

6 0

2 years ago