Answer:
We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution.
Explanation:
Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution. (MV) before dilution = (MV) after dilution. M before dilution = 1.0 M, V before dilution = ??? mL. M after dilution = 0.2 M, V after dilution = 100 mL. ∴ V before dilution = (MV) after dilution / M before dilution = (0.2 M)(100 mL) / (1.0 M) = 20.0 mL. So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution
D it is only physical and not chemical
Answer:
1 and 3 i think so i could be wrong
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.
