If 5.0mL of HC2H30 require 4.96mL of 0.9581 M NaOH to just consume the HC2H302, what is the

¿Cuál es la cantidad de electrones (e-) de Níquel si tiene una masa atómica de 58.6 y un número

daser333 [38]

Answer:

Hi do we translate a this

Explanation:

4 0

2 years ago

How many moles of H2O form when 4.5 moles O2 reacts?

mart [117]

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.

5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2

Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.

4 0

2 years ago

. Sulfur dioxide can be produced in the laboratory by the reaction of hydrochloric acid and a sulfite salt such as sodium sulfit

shutvik [7]

Answer:

Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g

Explanation:

SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.

the balanced chemical equation is as follows

Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O

Moles of Na₂SO₃ = $\frac{Mass}{Molecular mass} =\frac{25}{126} = 0.198$

Moles of HCl = $\frac{mass}{molecular mass}=\frac{22}{36.5}= 0.6$

using mole ratio method to find limiting reagent

For sodium sulfite $\frac{mole}{stoichiometry} = \frac{0.198}{1}= 0.198$

for HCl $\frac{mole}{stoichiometry} = \frac{0.6}{2}= 0.3$

since sodium sulfite is limiting reactant for above chemical reaction

1 mole of Na₂SO₃ produce 1 mole of SO₂

0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂

∴ Mass of SO₂ produce = mole x molar mass of SO₂

= 0.198 x 64

= 12.672 g

8 0

2 years ago

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

2 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

2 years ago