Fusion............................................................................
Answer:
60g of LiNO3 are required
Explanation:
<em>Calculate the amount of solute, in grams or milliliters, to prepare...</em>
<em />
The percentage mass/volume %(m/v) is defined as the mass in grams of the solute in 100mL of solution
To solve this question we must know that in a 37.5% (m/v) of LiNO3 you have 37.5g of LiNO3 in 100mL of solution.
Thus, to prepare 160mL are required:
160mL * (37.5g / 100mL) =
<h3>60g of LiNO3 are required</h3>
Answer:
Empirical CHO
molecular C4H4O4
Explanation:
From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.
This means : O = 100 - 41.39 - 3.47 = 55.14%
Next is to divide the percentage compositions by their atomic masses.
C = 41.39/12 = 3.45
O = 55.14/16 = 3.45
H = 3.47/1 = 3.47
Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.
Hence the empirical formula of Maleic acid is CHO
Now we go on to deduce the molecular formula.
To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.
Now we can see that 0.378mole = 43.8g
Then 1 mole = xg
x = (43.8*1)/0.378 = 115.87 = apprx 116
[CHO]n = 116
(12 + 1 + 16]n = 116
29n = 116
n = 116/29 = 4
The molecular formula is thus C4H4O4
Formula of sodium phosphate = Na₃PO₄
So, one mole of sodium phosphate molecules have 3 moles of Na atoms
Then, 1.6 moles of Na₃PO₄ will have: 1.6 × 3 = 4.8 moles of sodium atoms will be present