You can eliminate A, C, and D almost instantly if you know that engineers are construction workers. The answer is B.
Answer:
A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :
<u>2.67%</u>
Explanation:
Note : Look at the density of potassium nitrate in water if given in the question.
<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>
Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>
Density of potassium nitrate in water KNO3 = 2.11 g/mL

Density = 2.11 g/mL
Volume of solution = 86.4 mL



Mass of Solute = 4.87 g
Mass of Solution = 183.2 g
w/w% of the solute =


w/w%=2.67%
Answer:
Explanation:
Your strategy here will be to
use the chemical formula of carbon dioxide to find the number of molecules of
CO
2
that would contain that many atoms of oxygen
use Avogadro's constant to convert the number of molecules to moles of carbon dioxide
use the molar mass of carbon dioxide to convert the moles to grams
So, you know that one molecule of carbon dioxide contains
one atom of carbon,
1
×
C
two atoms of oxygen,
2
×
O
This means that the given number of atoms of oxygen would correspond to
4.8
⋅
10
22
atoms O
⋅
1 molecule CO
2
2
atoms O
=
2.4
⋅
10
22
molecules CO
2
Now, one mole of any molecular substance contains exactly
6.022
⋅
10
22
molecules of that substance -- this is known as Avogadro's constant.
In your case, the sample of carbon dioxide molecules contains
2.4
⋅
10
22
molecules CO
2
⋅
1 mole CO
2
6.022
⋅
10
23
molecules CO
2
=
0.03985 moles CO
2
Finally, carbon dioxide has a molar mass of
44.01 g mol
−
1
, which means that your sample will have a mass of
0.03985
moles CO
2
⋅
44.01 g
1
mole CO
2
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
1.8 g
a
a
∣
∣
−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the number of atoms of oxygen present in the sample.
Answer:
34 gram of FeO produced 8 gram of oxygen.
Explanation:
Given data:
Mass of FeO = 34 g
Mass of oxygen = ?
Solution;
Chemical equation:
2FeO → 2Fe + O₂
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 34 g /71.8 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of FeO with oxygen:
FeO : O₂
2 : 1
0.5 : 1/2 × 0.5 = 0.25
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g
So 34 gram of FeO produced 8 gram of oxygen.