Answer:
A(g)+B(g)⟶C(g) 2A(g)+2B(g)⟶5C(g)
A(s)+B(s)⟶C(g) 2A(g)+2B(g)⟶3C(g):
<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L
<u>Explanation:</u>
We are given:
Volume of mixture = 16 L
Percent of acid present = 30 %
Calculating the percentage of acid present in the mixture:

The mixture is made entirely of acid and water.
Volume of acid in the mixture = 4.8 L
Volume of water in the mixture = 16 - 4.8 = 11.2 L
Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L
Answer:
3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.
Explanation:
The molecular mass of P2O5 is
P2 = 2* 31 = 62
O5 = 5 *<u> 16 = 80</u>
Molecular Mass = 142
Set up a Proportion
142 grams P2O5 supplies 62 grams of phosphorus
x kg P2O5 supplies 1.69 kg of phosphorus
Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.
142/x = 62/1.69 Cross multiply
142 * 1.69 = 62x combine the left
239.98 = 62x Divide by 62
239.98/62 = x
3.89 kg of P2O5 must be used.