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Answer:
Cr₂S₃
Explanation:
From the question given above, the following data were obtained:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Empirical formula =?
Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Mass of sulphur (S) =?
Mass of S = (Mass of chromium sulfide) – (Mass of Cr)
Mass of S = 1.2888 – 0.67
Mass of S = 0.6188 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Cr = 0.67 g
Mass of S = 0.6188 g
Divide by their molar mass
Cr = 0.67 / 52 = 0.013
S = 0.6188 / 32 = 0.019
Divide by the smallest
Cr = 0.013 / 0.013 = 1
S = 0.019 / 0.013 = 1.46
Multiply by 2 to express in whole number
Cr = 1 × 2 = 2
S = 1.46 × 2 = 3
Therefore, the empirical formula of the compound is Cr₂S₃
Answer:
atoms of C? 2.25 x 1022 atoms C x 1 mole C = 0.037 mol C
Explanation:
How many moles of C are in 2.25 x 1022 atoms of C?
The answer to question 8 is
Ethanol's Molar Heat of Combustion. ... Heats of combustion are quoted as positive numbers while the enthalpy changes of combustion reactions (ΔH) are quoted as negative numbers, as combustion reactions are always exothermic. Heats of combustion are typically stated in kilojoules per mole (kJ/mol. or kJ mol. -1).
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Answer: 10 always love yourself
Explanation: