Answer:
147 g
Explanation:
<em>The number of moles of a substance is the ratio of the mass of the substance and the molar mass of that substance.</em> Mathematically,
mole = mass of substance/molar mass
In this case, Nick needs 3.5 moles of NaF, the molar mass of NaF is calculated as:
23 + 19 = 42 g/mol (Note: Na = 23, F = 19)
Hence,
mass of 3.5 mole NaF = mole x molar mass
= 3.5 x 42 = 147 g.
<em>Hence, Nick would need to measure out </em><em>147 g</em><em> NaF using a suitable weighing balance.</em>
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;

= 0.8
The rate-out
= 
= 
We can say that:

where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12

Integration of the above linear equation =

so we have:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places
Answer:
4.43 g of Oxygen
Explanation:
As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;
2 Moles of Aluminium
3 Moles of Sulfur
12 Moles of Oxygen
Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,
342.15 g ( 1 mole) of Al₂(SO₄)₃ contains = 192 g (12 mole) of O
So,
7.9 g of Al₂(SO₄)₃ will contain = X g of O
Solving for X,
X = (7.9 g × 192 g) ÷ 342.15 g
X = 4.43 g of Oxygen
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