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Nimfa-mama [501]
4 years ago
5

Determine the amount of energy (heat) in joules required to raise the temperature of 5 kg of iron from 50°C to 200°C.

Chemistry
1 answer:
larisa [96]4 years ago
4 0

333,000 Joules is the amount of energy (heat) in joules required to raise the temperature of 5 kg of iron from 50°C to 200°C.

Explanation:

Data given:

mass of iron 5 Kg or 5000

initial temperature = 50 degree centigrade

final temperature = 200 degrees centigrade

change in temperature (Δ T= 200 -50 degrees centigrade)

                                         = 150 °C

cp  (specific heat capacity of iron) = 0.444j/gram C

q (heat supplied) = ?

applying the formula,

q=mcΔT

putting the values in the equation:

q = 5000 X 0.444 X150

 q  = 333,000 Joules of energy.

The heat required 333,000 Joules of energy is required.

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Answer: The force of attraction occurring between two masses.

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How many grams of chloride are found in 72.03mg of magnesium chlorate?
olga_2 [115]

Answer:

26.73 mg.

Explanation:

  • Firstly, we can calculate the no. of moles of magnesium chlorate (Mg(ClO₃)₂):

no. of moles of magnesium chlorate (Mg(ClO₃)₂) = mass/molar mass = (72.03 mg)/(191.21 g/mol) = 0.377 mmol.

<em>Every 1.0 mole of magnesium chlorate (Mg(ClO₃)₂) contains 2.0 moles of Cl.</em>

<em></em>

∴ The no. of moles of Cl in magnesium chlorate (Mg(ClO₃)₂) = 2(0.377 mmol) = 0.754 mmol.

∴ The mass of Cl are found in 72.03 mg of magnesium chlorate (Mg(ClO₃)₂) = (no. of moles of Cl)(atomic mass of Cl) = (0.754 mmol)(35.453 g/mol) = 26.73 mg.

4 0
3 years ago
Describe the structure of atoms, including locations and charges of protons, neutrons, and electrons
pantera1 [17]

In an atom, the protons, which have a positive charge, and the neutrons, which have a neutral charge, are located in the nucleus. The electrons, which have a negative charge, orbit the nucleus of the atom.

3 0
3 years ago
What is 50% of 90??​
Alekssandra [29.7K]

Answer:

45

Explanation:

50% = 1/2

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3 years ago
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A solid mixture consists of 47.6g of KNO3 (potassium nitrate) and 8.4g of K2SO4 (potassium sulfate). The mixture is added to 130
IgorLugansk [536]

<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.

<u>Explanation:</u>

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be \frac{7.4}{100}\times 130=9.62g

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

7 0
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