The combustion of methane, CH4, releases 890.4 kJ/mol. That is, when one mole of methane is burned, 890.4 kJ are given off to the surroundings. This means that the products have 890.4 kJ less than the reactants.
Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Answer:
21.02moles of KBr
Explanation:
Parameters given:
Number of moles BaBr₂ = 10.51moles
Complete reaction equation:
BaBr₂ + K₂SO₄ → KBr + BaSO₄
Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:
The balanced equation is:
BaBr₂ + K₂SO₄ → 2KBr + BaSO₄
From the equation:
1 mole of BaBr₂ produces 2 moles of KBr
∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr
Answer: The transition elements are in the d-block, and in the d-orbital have valence electrons. They can form several states of oxidation and contain different ions. Inner transition elements are in the f-block, and in the f-orbital have valence electrons.
Explanation:
