Answer:
CaCO₃(s) => CaO(s) + CO₂(g) ... GpIIA Decomp
Explanation:
Metallic Carbonates decompose into a metallic oxide and carbon dioxide.
Examples:
Na₂CO₃(s) => Na₂O(s) + CO₂(g) ... GpIA Decomp
MgCO₃(s) => MgO(s) + CO₂(g) ... GpIIA Decomp
Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
Learn more about Partial pressure here:
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The answer is 7.33 g.
<span>To calculate this, we will use the the ideal gas law:
PV = nRT
where
P - pressure of the gas,
V - volume of the gas,
n - amount of substance of gas,
R - gas constant,
T - temperature of the gas.</span>
Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:
PV = RTm/M
It is given:
P = 0.98 atm
V = 10.2 l
T = 26°C = 299.15 K
R = 0.082 l atm/Kmol (gas constant)
M (H2O) = 2Ar(H) + Ar(O) = 2*1 + 16 = 2 + 16 = 18g
m = ?
Since PV = RTm/M, then:
m = PVM/RT
m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g
Answer:
treating everyone fairly
Explanation:
making or treating every fair as possible
