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
The total pressure = 1.402 atm
<u><em>calculation</em></u>
Total pressure = partial pressure of gas A + partial pressure of gas B + partial pressure of third gas
partial pressure of gas A= 0.205 atm
Partial pressure of gas B =0.658 atm
partial pressure for third gas is calculated using ideal gas equation
that is PV=nRT where,
p(pressure)=? atm
V(volume) = 8.65 L
n(moles)= 0.200 moles
R(gas constant)=0.0821 L.atm/mol.k
T(temperature) = 11°c into kelvin =11+273 =284 k
make p the subject of the formula by diving both side by V
p =nRT/v
p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm
Total pressure is therefore = 0.205 atm +0.658 atm +0.539 atm
=1.402 atm
Answer:
OXYGEN
I don't know what to say for the 20 character limit
Answer:
Explanation:
Hello!
In this case, when solid calcium carbonate, CaCO3 (s), is decomposed by the action of thermal energy (heat), solid calcium oxide, CaO (s) and carbon dioxide gas, CO2 (g) are yielded via the following reaction:
However, since calcium carbonate is solid as well as calcium oxide and carbon dioxide is given off as a gas, we write:
Which also balanced.
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