-log (1×10^-12) is how you calculate the pOH which in this case is 12
Answer: The empirical formula is 
.
Explanation:
Mass of C = 1.71 g
Mass of H = 0.287 g
Step 1 : convert given masses into moles.
Moles of C = 
Moles of H = 
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
The ratio of C: H = 1: 2
Hence the empirical formula is .
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
reproduction
Explanation:
reproduction, process by which organisms replicate themselves
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
<h3>
Molecular mass of potassium carbonate</h3>
The molecular mass of potassium carbonate, K₂CO₃ is calculated as follows;
M = K₂CO₃
M = (39 x 2) + (12) + (16 x 3)
M = 138 g
mass of carbon in potassium carbonate, K₂CO₃ is = 12 g
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is calculated as follows;
138 g ------------ 12 g of carbon
2.25 g ------------ ?
= (2.25 x 12) / 138
= 0.196 g
Thus, the mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
Learn more about potassium carbonate here: brainly.com/question/27514966
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