Lipids that are liquid at room temperature are known as oils.
Answer:
5.25g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is shown below:
Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O
From the balanced equation above,
8 moles of HF reacted to produce 2 moles of NaF.
Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.
Next, we shall convert 0.125 mole of NaF to grams.
This is illustrated below:
Mole of NaF = 0.125 mole
Molar mass of NaF = 23 + 19 = 42g/mol
Mass of NaF =..?
Mass = mole x molar mass
Mass of NaF = 0.125 x 42
Mass of NaF = 5.25g
Therefore, 5.25g of NaF is produced from the reaction.
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces. Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.
I hope this helps. Let me know if anything is unclear.
