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SCORPION-xisa [38]
3 years ago
5

What happens when carbon dioxide at 74°C is placed in thermal contact with water at 14°C? Energy leaves the carbon dioxide and e

nters the water. The water gets hotter than the carbon dioxide. The temperatures of the two substances equalize. Both the water and the carbon dioxide decrease in temperature. The water stays colder than the carbon dioxide.
Chemistry
1 answer:
qaws [65]3 years ago
5 0
<h3><u>Answer;</u></h3>

The temperatures of the two substances equalize.

<h3><u>Explanation;</u></h3>
  • When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same.
  • <em><u>The amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite. </u></em>
  • <em><u>In line with the law of conservation of energy, the amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. Therefore the amount of heat lost by carbon dioxide is equal to the amount of heat gained by water.</u></em>
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14.285 % is the answer maybe but I am not sure
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The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of
Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA  -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8



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What is the empirical formula of a compound containing 40.0% sulfur and 60.0% oxygen by mass?
gizmo_the_mogwai [7]

in order to determine empirical formula we have to determine the mole ratio of the given elements

Let the total mass of the compound is 100g

as given that the compound has 40% sulfur , so mass of sulfur = 40g

as given that the compound has 60% oxygen, so mass of oxygen = 60g

let us calculate the moles of each element

Moles of sulfur = mass / atomic mass = 40 / 32 = 1.25

moles of oxygen = mass / atomic mass = 60/ 16 = 3.75

In order to get simple ratio of moles we will divide both the moles with least number of moles which is 1.25

moles of sulfur = 1.25 / 1.25 = 1

moles of oxygen = 3.75 /1.25 = 3

So empirical formula will be SO₃

3 0
3 years ago
Ammonia reacts with sulfuric acid to produce the important fertilizer, ammonium hydrogen sulfate.
Liula [17]

Answer:

404.8g of (NH4)HSO4 is produced.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Temperature (T) = 10°C = 10°C + 273 = 283K

Pressure (P) = 110KPa = 110/101.325 = 1.09atm

Volume (V) = 75L

Step 2:

Determination of the number of mole of ammonia, NH3.

The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.09 x 75 = n x 0.0821 x 283

Divide both side by 0.0821 x 283

n = (1.09 x 75) /(0.0821 x 283)

n = 3.52 moles

Step 3:

Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NH3 + H2SO4 —> (NH4)HSO4

From the balanced equation above,

1 mole of NH3 produced 1 mole of (NH4)HSO4

Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.

Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.

Step 4:

Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:

Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol

Number of mole of (NH4)HSO4 = 3.52 moles

Mass of (NH4)HSO4 =..?

Mass = mole x molar Mass

Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g

Therefore, 404.8g of (NH4)HSO4 is produced.

5 0
2 years ago
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