Answer:
50 g of K₂CO₃ are needed
Explanation:
How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?
We analyse data:
500 g is the mass of the solution we want
10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution
Therefore we can solve this, by a rule of three:
In 100 g of solution we have 10 g of K₂CO₃
In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃
Answer:
It's an animal that hunts anothe animal in way to eat it
Explanation:
Answer:
1. 504.8 g Al(NO3)3
2. 14.3 moles O2
Explanation:
1.
g = moles x molar mass = 2.37 x 212.996 = 504.8 g Al(NO3)3
MM Al(NO3)3 = 212.996 g/mol
2.
moles = mass : molar mass = 456.89: 32 = 14.3 moles O2
Answer:
The final pressure in the container at 0°C is 2.49 atm
Explanation:
We apply the Ideal Gases law to know the global pressure.
We need to know, the moles of each:
P He . V He = moles of He . R . 273K
(1atm . 4L) / R . 273K = moles of He → 0.178 moles
P N₂ . V N₂ = moles of N₂ . R . 273K
(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles
P Ar . V Ar = moles of Ar . R . 273K
(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles
Total moles: 0.892 moles
P . 8L = 0.892 mol . R . 273K
P = ( 0.892 . R . 273K) / 8L = 2.49 atm
R = 0.082 L.atm/mol.K
Answer:
c. Waxing crescent
- I hope this helps have a great night
