In physics, power is the rate of doing work. It is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. In the SI system, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the steam engine.
Answer:
The core elements are hydrogen and helium.
Explanation:
The balanced chemical equation would be as follows:
<span>NaCl + AgNO3 -> NaNO3 + AgCl
We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl -----> consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3
0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced
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Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
Tin to Fluorine mass ratios:
1) For compound A:
38.5/12.3
= 3.13
2) For compound B:
56.5/36.2
= 1.56
The lowest whole number mass ratio is 2. It cannot be 1 because it is less than that required for compound B.
