Answer:
29.575%
Explanation:
Data provided:
Calories taken in daily diet = 2000
Recommended amount of fat = 65 grams
Average number of calories for fat = 9.1 calories / g
Thus,
Number of calories in the diet with average number of calories for fat
= Recommended amount of fat × Average number of calories for fat
= 65 × 9.1
= 591.5 calories
Therefore,
the percentage of calories in his diet supplied = ( 591.5 / 2000 ) × 100
= 29.575%
Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30
The concentration of diluted solution is 0.756M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 18.9 mL
Molarity of stock solution (M1) = 10 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
10 × 18.9 = M2 ×250
189 = M2 × 250
Divide both side by 100
M2 = 189 / 250
M2 = 0.756 M
Therefore, the molarity of the diluted solution is 0.756 M.
Thus the concluded that concentration of the dilute acid is 0.756 M.
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Answer: 4.4 x 10^-7
Explanation:
The dissociation equation for this reaction is:
MgCO3 (s) → Mg+2 (aq) + CO3-2 (aq)
(Here 0.08 >>> x )
So the solubility MgCO₃ in a solution that containing 0.080 M Mg²⁺ is 4.4 x 10^-7
