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Paladinen [302]
3 years ago
12

Potassium chlorate is a mean little chemical. It likes to decompose rather violently, and when it does, it decomposes into two d

ifferent products--potasssium chloride and a gas. What is this gas produced in the reaction?
Chemistry
1 answer:
WARRIOR [948]3 years ago
6 0

Answer:

Oxygen gas

Explanation:

Potassium chlorate KClO3 decomposes thermally to form potassium chloride KCl and oxygen gas is given off. This is in the presence of a catalyst, manganese iv oxide.

Hence, what is being said in essence is that when potassium chloride is heated in the presence of manganese iv oxide catalyst, oxygen is given off as the gaseous product with the formation of potassium chloride.

The chemical equation for the reaction is :

2KClO3 ——-> 2KCl + 3O2

This procedure is one of most important ways through which oxygen can be produced in the laboratory and can be used as a source of oxygen is industrial processes if on a large scale

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
2 years ago
Which option draws the correct conclusion from the following case study?
trasher [3.6K]

Answer:

I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong

Explanation:

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Metallic bonding is...
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Answer:

d an attraction between positive ions and electrons

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What are the 7 methods of separating mixtures?​
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Methods Of Separating Mixtures

Handpicking.

Threshing.

Winnowing.

Sieving.

Evaporation.

Distillation.

Filtration or Sedimentation.

Separating Funnel.

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3 years ago
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Which is the metric standard for measuring energy? Which unit is used for specific heat capacity? If you wanted to compare the a
DENIUS [597]

1. Which is the metric standard for measuring energy?

The <u>metric standard for measuring of energy </u>defined by the International System of Units is the joule (J), which is defined as the work done by a force of a newton in a displacement of one meter in the direction of force. So,

1 J = 1 N m = 1 kg·m²/s²

Calorie is also frequently used in scientific and technological applications. Calorie is a <u>unit of thermal energy that is equivalent to the amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius</u>.

1 cal = 4,184 J

2. Which unit is used for specific heat capacity?

The specific heat capacity (c) is a physical quantity that is defined as the <u>amount of heat (</u><u>q</u><u>) that must be supplied to the mass unit of a thermodynamic substance or system to raise its temperature by one unit.</u> So,

c = q / m ΔT

where m is the mass of the substance and ΔT is the temperature increase.

In this way, as heat is a form of energy, the International System of Units expresses the specific heat in <u>joules per kilogram and per kelvin</u> (J kg⁻¹ K⁻¹). Another common unit, not belonging to the SI, is the <u>calorie per gram and per degree centigrade</u> (cal g⁻¹ ° C⁻¹).

3. If you wanted to compare the abilities of olive oil and peanut oil to gain or lose thermal energy, which unit would you use?

You should use units of specific heat capacity (J kg⁻¹ K⁻¹) since, as mentioned above, this is a physical quantity that measures the amount of heat that must be supplied to an specific mass of a substance or system to raise its temperature.

Heat is a thermal energy, so <u>by using heat capacity units you can compare the ability of</u><u> </u><u>olive oil and peanut oil to gain or lose thermal energy by varying its temperature.</u>

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