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3 years ago

Which statement is true of any chemical reaction?

Chemistry

1 answer:

klemol [59]3 years ago

4 0

According to the <em>Law of Conservation of Mass</em>,

The mass of the products in a chemical reaction must equal the mass of the reactants.

∴ D is the Answer

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TIMED PLS HELP AND WILL GIVE BRAINLIST

Strike441 [17]

Answer:

19.4 g of alum, will be its theoretical yield

Explanation:

The reaction is:

2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O

Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles

Ratio is 4:2

4 moles of sulfuric acid can make 2 moles of alum

By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.

We convert moles to mass:

Molar mass of alum is: 473.52 g/mol.

0.041085 moles . 473.52 g/mol = 19.4 g

6 0

2 years ago

The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens

poizon [28]

Answer:

<u>It increases by a factor of four</u>

Explanation:

Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.

pV = K.......(1)

pV = constant

Charles law : The volume of the gas is directly proportional to temperature at constant pressure.

V = KT

or V/T = K = constant ....(2)

Applying equation (1) and (2)

$\frac{PV}{T}=K$

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

According to question ,

T2 = 4 (T1)

V2 = V1

Put the value of T2 and V2 , The P2 can be calculated,

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{1}}{4T_{1}}$

V1 and V1 cancel each other

T1 and T1 cancel each other

We get,

$P1=\frac{P2}{4}$

P2 = 4 P1

So pressure increased by the factor of four

5 0

3 years ago

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.