Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
Answer:
Explanation:
If a substance is a limiting reactants then the chemical reaction will not last a long time because the reactant has a set limit it will stop reacting with the second reactant. Hope this helped :)
Answer:
c : 13%
Explanation:
Data Give:
Experimental density of vanadium = 6.9 g/cm³
percent error = ?
Solution:
Formula used to calculate % error
% error = [experimental value -accepted value/accepted value] x 100
The reported accepted density value for vanadium = 6.11 g/cm³
Put value in the above equation
% error = [ 6.9 - 6.11 / 6.11 ] x 100
% error = [ 0.79 / 6.11 ] x 100
% error = [ 0.129] x 100
% error = 12.9
Round to the 2 significant figure
% error = 13 %
So, option c is correct
*** 2 ***
<span>if we assume volume NaCl + volume H2O = volume H2O.. i.e.. NaCl does not effect volume </span>
<span>therefore.. the units of.. </span>
<span>.. M = moles NaCl / L solution ≈ moles NaCl / L H2O </span>
<span>.. density = grams NaCl / L solution ≈ grams NaCl / L H2O </span>
<span>again.. that is our assumption </span>
<span>so we can readily see that </span>
<span>.. M = (1 mol NaCl / ___g NaCl) x (__g NaCl / L H2O) + 0 </span>
<span>ie.. </span>
<span>.. M = (1 mol NaCl / 58.5g NaCl) x density solution + 0 </span>
<span>so.. we would expect.. </span>
<span>.. m = 0.01709 mol / g </span>
<span>.. b = 0 </span>
Answer:
Only white phosphorus is stored under water. White phosphorus spontaneously reacts with oxygen in the air to burst into flame to form phosphorus pentoxide
Explanation:
