Answer : 50.69 mg of ascorbic acid does not meet the daily requirement.
Solution : Given,
Molar mass of Ascorbic acid = 176 g/mole
Moles of Ascorbic acid = 
Formula used :

or, 
Now put all the given values in this formula, we get the mass of ascorbic acid.

Conversion : 
As per question, a healthy adult’s daily requirement of vitamin C is 70-90 mg. But calculate mass of vitamin C is 50.69 mg. So, 50.69 mg of ascorbic acid does not meet the daily requirement.
Answer:
1.09 moles of NaOH
Explanation:
First of all, to calculate moles, you need to find the molar mass of NaOH.
Let us first find the molar mass of NaOH then.
Na = 23.0 amu
O = 16.0
H = 1.0
They are 1 nitrogen atom, 1 oxygen atom, and one hydrogen atom.
So do this.
23.0(1) + 16.0(1) + 1.0(1) = 40 g/mol.
Now use dimensional analysis to show your work
43.5 g of NaOH * 1 mol of NaOH / 40 g/mol of NaOH
The grams cancel out.
43.5 / 40.0 = 1.0875
Use sig figs and round the answer to the nearest hundredths place.
1.0875 = 1.09
So the final answer is 1.09 moles of NaOH
Hope it helped!
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
<em />
Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
<em />
Suppose X (in unit M) be the required maximum concentration of the Pb(NO₃)₂ solution added after mixing with 100 ml of 6.5 x 10⁻² M NaCl, the Pb²⁺ concentration is:
[Pb²⁺] = (X*100) / 200 = 0.5 X
The concentration of Cl⁻ is:
[Cl⁻] = (100 / 200) * (6.5 x 10⁻² M) = 0.0325 M
So:
Ksp = 2.00 x 10⁻⁵ = [Pb²⁺] [Cl⁻]² = (0.5X) * (0.0325)²
X = (2.00 x 10⁻⁵) / (5.28 x 10⁻⁴) = 0.038 M