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3 years ago

Read the following reactions.

Chemistry

1 answer:

garik1379 [7]3 years ago

5 0

Answer:

Only Reaction 1

Explanation:

In reaction 1, there is a change in state from solid to liquid. Hence, there is an increase in number of ways particles and their energies could be arranged. As a result, entropy increases.

In reaction 2, there is a decrease in amount of gas particles (4 mol to 2 mol). Hence there is a decrease in the number of ways particles and their energies could be arranged. As a result entropy decreases

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How does a metric prefix modify a unit?

Daniel [21]

Answer:Locate the row corresponding to known unit of torque along the left of the table. Multiply by the factor under the column for the desired units. For example, to convert 2 oz-in torque to n-m, locate oz-in row at table left. Locate 7.062 x 10-3 at intersection of desired n-m units column. Multiply 2 oz-in x (7.062 x 10-3 ) = 14.12 x 10-3 n-m.

Converting between units is easy if you have a set of equivalencies to work with. Suppose we wanted to convert an energy quantity of 2500 calories into watt-hours. What we would need to do is find a set of equivalent figures for those units. In our reference here, we see that 251.996 calories is physically equal to 0.293071 watt hour. To convert from calories into watt-hours, we must form a “unity fraction” with these physically equal figures (a fraction composed of different figures and different units, the numerator and denominator being physically equal to one another), placing the desired unit in the numerator and the initial unit in the denominator, and then multiply our initial value of calories by that fraction.

Explanation:

Since both terms of the “unity fraction” are physically equal to one another, the fraction as a whole has a physical value of 1, and so does not change the true value of any figure when multiplied by it. When units are canceled, however, there will be a change in units. For example, 2500 calories multiplied by the unity fraction of (0.293071 w-hr / 251.996 cal) = 2.9075 watt-hours.

7 0

3 years ago

Chose the mathematical formula expressing work.

Gelneren [198K]

W = AB x F x Cos < AB, F
or just W= AB x F for short

7 0

2 years ago

What colour is litmus in a solution of pH 4?

schepotkina [342]

Answer:

in solution of PH4 it is red

in water it is blue

6 0

2 years ago

Describe endothermic vs. exothermic reactions in terms of product formation. (hint: what happens when products are formed?)

shusha [124]

Answer:

Endothermic reaction chemical equation

Reactnt A + Reactant B + Heat (energy) ⇒ Products

Exothermic reaction chemical equation

Reactnt A + Reactant B ⇒ Products + Heat (energy)

Explanation:

Endothermic Reaction

An endothermic reaction is a reaction that reaction that requires heat before it would take place resulting in the absorption of heat from the surrounding that can be sensed by the coolness of the reacting system

An example of an endothermic reaction is a chemical cold pack that becomes cold when the chemical and water inside it reacts

Exothermic Reaction

An exothermic reaction is one that rekeases energy to the surroundings when it takes place. This is as a result of the fact that the combined heat energy of the reactants is more than the chemical heat energy of the products. An example of an exothermic reaction is a burning candle

5 0

2 years ago

Rank the following fertilizers in decreasing order of mass percentage of nitrogen:

charle [14.2K]

<h3>Answer:</h3>

NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄ = 132.06 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age = 21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age = 21.18 %

In KNO₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of KNO₃ = 101.10 g.mol⁻¹

Mass %age = Mass of N / M.Mass of KNO₃ × 100

Mass %age = 14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age = 13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄ = 115.03 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age = 14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age = 12.17 %

In NH₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of NH₃ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₃ × 100

Mass %age = 14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age = 82.20 %

In NH₄NO₃:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of NH₄NO₃ = 80.04 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age = 28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age = 34.98 %

5 0

3 years ago