Which of the following is the best example of kinetic energy being transformed into potential energy?

kicyunya [14]

The rate of the reaction is measurable quantity that refers to the amount or how much is are chemical substances reagents used up or converted into the product over some period of time.

Rate = change in the amount/time.

This can indirectly be observed through many ways, such as the volume of gas given off if the byproduct is a gas being produced, the colour of the solution etc.

4 0

3 years ago

What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0

3 years ago

What is the concentration in g/L of a solution containing 10 g of solute in 2 L of solution

alekssr [168]

The concentration is 5 g/L.

Concentration = mass/volume = 10 g/2 L = 5 g/L

6 0

3 years ago

How many grams of acetylene are produced by adding water to 5.00 g CaC2?

Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

5 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