Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Explanation:
The given data is as follows.
= 10 mM = 
M
= 750 ml, 
= 5 ml
= ?
Therefore, calculate the molarity of given NaCl stock as follows.
= 1.5 M
Thus, we can conclude that molarity of given NaCl stock is 1.5 M.
<span>a. It melts at 1455oC I know this is correct I need One more</span>
Answer:
The answer to your question is 3% H2SO4 solution
Explanation:
Data
Concentration 2 = C₂ = ?
Concentration 1 = C₁ = 15 %
Volume 1 = V₁ = 50 ml
Volume 2 = V₂ = 250 ml
Formula
C₁V₁ = C₂V₂
Solve for C₂
C₂ = C₁V₁ / V₂
Substitution
C₂ = (15)(50) / 250
Simplification and result
C₂ = 3 %
-2 cuz 20 negative minus 18 positive is -2
