Which one of the following is not an acid? a.Option 1 b.HOOCCOOH c.H2SO4 d.NaCl

How many moles of SnF₂ will be produced along with 48 grams of H₂? *

KIM [24]

Answer: 24 moles of $SnF_2$ are produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

${\text{Moles of} H_2}=\frac{48g}{2g/mol}=24moles$

$Sn+2HF\rightarrow SnF_2+H_2$

According to stoichiometry :

1 mole of $H_2$ is accompanied with = 1 mole of $SnF_2$

Thus 24 moles of $H_2$ is accompanied with = $\frac{1}{1}\times 24=24moles$ of $SnF_2$

Thus 24 moles of $SnF_2$ are produced.

8 0

3 years ago

Why is hydgron content with only 2 electrons instead of 8 like atoms in periods greater than 1?

Cerrena [4.2K]

Because the sublevels 1s only has one shape ,and it only contains 2 elections...

5 0

3 years ago

ΔH for the formation of CuCl2 from its elements is -220.1 kJ/mol. How many kJ are associated with the formation of 0.30 mole of

otez555 [7]

0.3 * - 220.1 = - 66.03kJ

3 0

3 years ago

Read 2 more answers

A student was given a piece of metal with a mass of 85.0 g. She placed it

Anastasy [175]

Answer:

22mL

Explanation:

6 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago