Answer:
280 g Al₂O₃
Explanation:
To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).
Molar Mass (Al₂O₃): 101.96 g/mol
2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole
The final temperature : 345 K
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Further explanation
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Given
475 cm³ initial volume
600 cm³ final volume
Required
The final temperature
Solution
At standard temperature and pressure , T = 273 K and 1 atm
Charles's Law :
When the gas pressure is kept constant, the gas volume is proportional to the temperature
V₁/T₁=V₂/T₂
Input the value :
T₂=(V₂T₁)/V₁
T₂=(600 x 273)/475
T₂=345 K
Answer:
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Explanation:
Answer:
Explanation:
Chlorine is the element of group 17 and third period. The atomic number of chlorine is 17 and the symbol of the element is Cl.
The electronic configuration of the element chlorine is:-
Chloride ion is formed when chlorine atom gain one more electron. So, the ground-state electron configuration for the chloride ion is:-
Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm