Answer:
1045
Answer: 1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.
Hello!
When finding the chemical formula of a compound, we will need to find the charges of each element/bond.
Looking at our period table, sodium has a +1 charge, written as Na 1+, and sulfate has a charge of -2, and it is written as SO4 2-.
Now, we need to make the charges equivalent. To do this, we need to "criss-cross" the charges. This means that sodium will need to additional atoms to make the charges equal, and sulfate will need one.
Therefore, the chemical formula for sodium sulfate is: Na2SO4.
First, find how many grams are in 1 mole of water.
For a hydrogen atom, there is about 1 gram per mole. For an oxygen atom, there are about 16 grams per mole.
In H2O, there are two hydrogen atoms and one oxygen atom. This means there are 18 grams in one mole of water. Multiply the mass in one mole by your number of moles.
18 x 11.8 = 212.4 grams
You have 212.4 grams of water.
The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
to learn more about energy
brainly.com/question/7185299
#SPJ4
<h3>
Answer:</h3>
812 kPa
<h3>
Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,
At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;
Replacing the known variables with the values;
= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa
