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klio [65]
3 years ago
11

What is the heat of fusion of water in calories per mole?​

Chemistry
2 answers:
Fed [463]3 years ago
7 0

Answer:The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,230 joules

Explanation: i really hope this helps but im not sureif this is exactly what your looking for :)

Mumz [18]3 years ago
6 0

Answer:

<u><em>79.72 Calories</em></u>

Explanation:

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Which is a sign of a chemical change?
Igoryamba
The correct answer for the question it is option  

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7 0
3 years ago
How many atoms are in 8.28 moles of aluminum?
Lilit [14]

Answer:

49.86 × 10²³  atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³  atoms of Al

8.28 mol×6.022 × 10²³  atoms / 1mol

49.86 × 10²³  atoms of Al

4 0
3 years ago
Please help this is so hard 1: Why is the sky blue?
kolezko [41]

Answer:

1. As white light passes through our atmosphere, tiny air molecules cause it to 'scatter'. Therefore, blue light is scattered more than red light and the sky appears blue during the day.

2. We can see the moon during the day for the same reason we see the moon at night. The surface of the moon is reflecting the sun's light into our eyes. When we see the moon during the day it's because the moon is in the right spot in the sky and it's reflecting enough light to be as bright, or brighter, than the sky.

3. At eight pounds per gallon, that's another 38,593,169,564 pounds of water in the air. Add it all up, and the total weight of our sky is 12,107,415,382,493,169,564 pounds, more than 12 quintillion pounds!

4. 5.972 × 10^24 kg

5. Four forces keep an airplane in the sky. They are lift, weight, thrust and drag. Lift pushes the airplane up.

6. Water is wet, in the sense of being a liquid which flows easily, because its viscosity is low, which is because its molecules are rather loosely joined together.

7. Rainbows are formed when sunlight is scattered from raindrops into the eyes of an observer. The lower the sun in the sky the more of an arc of a rainbow the viewer will see. Rain, fog or some other source of water droplets must be in front of the viewer.

8. Birds can sit on power lines and not get electric shocks because the electricity is always looking for a way to get to the ground. The birds are not touching the ground or anything in contact with the ground, so the electricity will stay in the power line.

Explanation:

4 0
3 years ago
Read 2 more answers
Determine the mass of a gold bar that has a density of 19.3 g/cm3 and is 4.72 cm high by 8.21 cm long by 3.98 cm deep.
TEA [102]
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
5 0
3 years ago
3) A child has a toy balloon with a volume of 1.8 liters. The temperature of the balloon when it was filled was 22° C and the pr
mestny [16]

This is a combined gas law problem, according to which

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P is the pressure of the gas, V is the volume of the gas, and T is the temperature of the gas, and the subscripts 1 and 2 correspond to the initial and final conditions of the gas. In this problem, we are given the initial pressure, volume, and temperature of the gas in the balloon:

P₁ = 1.0 atm

V₁ = 1.8 L

T₁ = 295.15 K (K = °C + 273.15).

Moreover, we are given the final pressure and temperature of the gas in the balloon.

P₂ = 0.86 atm

T₂ = 281.15 K.

What we want to find is the final volume, V₂, which we can obtain by rearranging the combined gas equation to solve for V₂:

V_2=\frac{P_1V_1T_2}{T_1P_2} = \frac{(1.0 \text{ atm})(1.8 \text{ L})(281.15 \text{ K})}{(295.15 \text{ K})(0.86 \text{ atm})} \\ V_2 = 1.99 \text{ L}

This answer has three significant figures. However, the question as written would warrant an answer that comprises one significant figure (as 8 °C has only one sig fig). In that case, the answer would be 2 L. If the answer is to be given to two significant figures, the volume would then be 2.0 L.

8 0
3 years ago
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