A group 2 metal carbonate has a mass of 84 g/mol. Identify the group 2 metal X using its chemical formula.
1 answer:
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles =
Number of moles = = 0.003mole
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Answer: -
405 mL
Explanation: -
Initial Pressure P 1 = 0.6 atm
Final Pressure P 2 = 0.2 atm
Initial volume V 1 = 135 mL
Using Boyle's Law,
At constant temperature,
P 1 x V 1 = P 2 x V 2
0.6 atm x 135 mL = 0.2 atm x V 2
Final volume = V 2 =
= 405 mL
Thus if a sample of gas has a volume of 135 ml at 0.600 atm. 405 mL would be the volume if the pressure is decreased to 0.200 atm while temperature is held constant.