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user100 [1]
3 years ago
11

How many particles would be found in a 1.224 g sample of K2O

Chemistry
2 answers:
irina [24]3 years ago
6 0

Answer:

7.8286×10²¹ particles.

Explanation:

First we need to calculate the total molar mass of the compound, in this case:

Potassium (K) = 39.1 g/mol × 2 =  78.2 +

Oxigen (O)     =     16 g/mol × 1  = <u>    16   </u>

                                Total(K₂O) = 94.2 g/mol

Then, we calculate the number of moles of the compound in the sample, this is done dividing de mass of the sample by the molar mass:

mol =\frac{1.224 g}{94.2 g/mol}

mol = 0.013 moles in our sample.

Finally, we calculate the total number of particles. The costant known as Avogadro number (6.022×10²³) is the number of particles or atoms contained in a mole of any substance. We need to multiply the number of moles by the Avogadro number.

particles = 0.013 mol × (6.022×10²³  particles/mol) = 7.8286×10²¹ particles.

lbvjy [14]3 years ago
4 0

Answer:

9.96*10^21

Explanation:

Molar mass of K2O=29*2+16

= 74g per mol

number of moles in the sample= 1.224/ 74

=0.1654

Number of particles in 1 mole=6.0221409*10^23

Number of particles= 0.01654*6.0221409*10^23

=9.96*10^21

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How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?
Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water  at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

6 0
3 years ago
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