In binary ionic compounds the name of the cation (Metal) is first, so that’s how you know.
I think it would be solubility but I’m not sure
The first step is to balance the equation:
<span>C3H8 + 5O2 ---> 3CO2 + 4H2O
Check the balance
element left side right side
C 3 3
H 8 4*2 = 8
O 5*2=10 3*2 + 4 = 10
Then you have the molar ratios:
3 mol C3H8 : 5 mol O2 : 3 mol CO2 : 4 mol H2O
Now you have 40 moles of O2 so you make the proportion:
40.0 mol O2 * [3 mol CO2 / 5 mol O2] = 24.0 mol CO2.
Answer: option D. 24.0 mol CO2
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There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
Answer: I think the answer is, 3.312177825e+24 atoms
Explanation: I had a problem similar to this, Hope this helps!
