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mrs_skeptik [129]
2 years ago
11

Which of the following will increase entropy in a reaction?

Chemistry
1 answer:
Lapatulllka [165]2 years ago
3 0

B. Heating up the reaction will increase the entropy of a reaction.

<h3>What is entropy?</h3>

Entropy is the measure of the degree of disorderliness of a system.

Entropy is also the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

S = ΔH/T

where;

  • S is entropy
  • ΔH is energy input
  • T is  temperature

Entropy increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules.

However, entropy increases as temperature increases. Thus, heating up the reaction will increase the entropy of a reaction.

Learn more about entropy here: brainly.com/question/6364271

#SPJ1

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Calculate how many grams of o2(g) can be produced from heating 87.4 grams of kclo3(s).
Rufina [12.5K]
<span>34.2 grams Lookup the atomic weights of the involved elements Atomic weight potassium = 39.0983 Atomic weight Chlorine = 35.453 Atomic weight Oxygen = 15.999 Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol The balanced equation for heating KClO3 is 2 KClO3 = 2 KCl + 3 O2 So 2 moles of KClO3 will break down into 3 moles of oxygen molecules. 0.713188188 mol / 2 * 3 = 1.069782282 mols So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be 1.069782282 * 2 * 15.999 = 34.23089345 grams Rounding the results to 3 significant figures gives 34.2 grams</span>
4 0
3 years ago
Trying to explain why a cactus needs little water to survive is an example of
zmey [24]
It’s a or b i think it’s b
6 0
3 years ago
What will happen if a peeled banana is put on a hotplate?
inysia [295]
It will melt or the molecules inside of it will get hot
8 0
3 years ago
Can someone help me with number 1 and 2 plz!
GalinKa [24]

Answer:

1) 0 N

2) 8 N

Explanation:

The net force is the sum of all of the forces acting on the object.

For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left.  If we define the right to be positive and the left to be negative, then the net force equals:

Fnet = 5N - 5N = 0 N

Therefore, the net force in question 1 is 0 N.

For question 2, the process is very similar.  We want to find the sum of the forces acting on the object.  In this case, there are forces of 3 N and 5 N acting to the right.

Fnet = 3 N + 5 N = 8 N

Therefore, the net force in question 2 is 8 N.

Hope this helps!

3 0
3 years ago
a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what
jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

Where:

P - Pressure, measured in kilopascals.

V - Volume, measured in liters.

n - Molar quantity, measured in moles.

T - Temperature, measured in Kelvin.

R_{u} - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}} (2)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in liters.

T_{1}, T_{2} - Initial and final temperature, measured in Kelvin.

If we know that P_{1} = 121\,kPa, P_{2} = 202\,kPa, V_{1} = 2.7\,L, T_{1} = 288\,K and T_{2} = 303\,K, the final volume of the gas is:

V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}

V_{2} = 1.702\,L

The gas will occupy a volume of 1.702 liters.

6 0
3 years ago
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