Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %
Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
We can solve this without a concrete formula through dimensional analysis. This works by manipulating the units such that you end up with the unit of the final answer. Manipulate them by cancelling units that appear both in the numerator and denominator side. As a result, we must be left with the units of g. The current in A or amperes is equivalent to amount of Coulombs per second. Since this involves Coulombs, we will use the Faraday's constant which is 96,500 C/mol electron. The reaction is:
Cr³⁺(aq) + 3e⁻ --> Cr(s)
This means that for every 3 moles of electron transferred, 1 mole of Chromium metal is plated. The molar mass of Cr: 52 g/mol. The solution is as follows:
Mass of Chromium metal = (8 C/s)(60 s/1 min)(160 min)(1 mol e⁻/96,500 C)(1 mol Cr/3 mol e)(52 g/mol)
<em>Mass of Chromium metal = 13.79 g</em>
Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1
Knowing the number of valence electrons in one of the alien elements helps in identifying it because the number of valence electrons can help categorize the alien element. Similar elements have the same valence electrons and knowing the category of the element can help further analyze the element.
