Answer:
4Ba(CO3) -> 4BaO2 + 2CO2
Explanation:
I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.
Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide = 
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
Answer:
(NH4)2CO3 is the formula !
Answer:
Molarity= 1.69M
Explanation:
m= 14.8, Mm= 35, V= 0.25dm3, C= ?
Moles = m/M= C×V
Substitute and Simplify
m/M= C×V
14.8/35= C×0.25
C= 1.69M
