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brilliants [131]
3 years ago
11

How much heat is required to raise the temperature of 0.50 kg of liquid water from 15℃ to 90℃?

Chemistry
1 answer:
jolli1 [7]3 years ago
4 0
This will require 266.9k j of heat energy my explanation for this is to calculate the energy required to raise the temperature of any given substance here what you require the mass of the material,m the temperature change that occurs AT
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Why did Dalton think it was important to use his system of symbols for the chemical elements?
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The answer is : A good luck :)
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Greenhouse effect is warming due to
oee [108]

Answer:

This effect is caused by increased levels of carbon dioxide, chlorofluorocarbons and other gases in the air, many of them released by human activity. greenhouse effect The warming of Earth's atmosphere due to the buildup of heat-trapping gases, such as carbon dioxide and methane.

Explanation:

6 0
3 years ago
What do methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C3H6O3) all have in common?
tatyana61 [14]

Answer:

d

Explanation:

Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Therefore, a is true.

An empirical formula is the simplest ratio of atoms present in a compound. Therefore, C2H4O2 and C3H6O3, (if you simplified them like you would a fraction) would be CH2O. Therefore b is correct,

They also have the same % composition, with a ratio of 1 carbon : 2 hydrogen : 1 oxygen. Therefore, c is correct.

Since a, b and c are all correct, the answer is d, all of the above are true.

6 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
Please help 21 points
Novosadov [1.4K]

Answer:

You

Explanation:

Will have to fill in the graph organizer with a story

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