Answer:
28 g CO
Explanation:
First convert grams to moles.
1 mole C = 12.011 g (I'm just going to round to 12 for the sake of this problem)
12 g C • 
= 1 mol C
1 mol O = 15.996 g (I'm just going to round to 16)
16 g O • 
= 1 mol O
So the unbalanced equation is:
-> 
(the oxygen has a 2 subscript because it is part of HONClBrIF meaning when not in a compound these elements appear in pairs - called diatomic elements)
The balanced equation is:
-> 
However, carbon is the limiting reactant in this equation and two moles cannot react because only 12 g (1 mole) are present. Therefore, use the equation
-> .
1 mole of CO is formed, therefore 12 g + 16 g = 28 g CO.
If combined with oxygen. the reaction is usually combustion. in this case, it is the combustion of ethene
Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Answer:
n = 2 moles (1 sig-fig)
Explanation:
Using the Ideal Gas Law equation (PV = nRT), solve for n (= moles) and substitute data for ...
pressure = P(atm) = 100atm
volume =V(liters) = 50L
gas constant = R = 0.08206L·atm/mol·K
temperature = T(Kelvin) = °C + 273 = (35 + 273)K = 308K
PV = nRT => n = PV/RT = (100atm)(50L)/(0.08206L·atm/mol·K)(308K)
∴ n(moles) = 1.978moles ≅ 2 moles gas (1 sig-fig) per volume data (= 50L) that has only 1 sig-fig. (Rule => for multiplication & division computations round final answer to the measured data having the least number of sig-figs).
The best and most correct answer among the choices provided by your question is the second choice or letter B.
The iron cam is larger than the aluminum cam even if with the same size.
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