Answer:
m = 20.9 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:
Regards!
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
Learn more about heat transfer:
brainly.com/question/10286596
#SPJ1
Answer:- 
molecules.
Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.
It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.
In second step, the moles are converted to molecules on multiplying by Avogadro number.
Molar mass of 
= 12+4(79.9) = 331.6 g per mol
let's make the set up using dimensional analysis:
= molecules
So, there will be molecules in 250 grams of .
I would say diamond, because it consists only of carbon
