Answer:
1.67g H2CO3 are produced
Explanation:
Based on the reaction:
2NaHCO3 → Na2CO3 + H2CO3
<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>
To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:
<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>
2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3
<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>
0.02698 moles * (62.03g/mol) =
<h3>1.67g H2CO3 are produced</h3>
Molecular weight of AgBr = 187.7
moles of Ag =

moles of Br = moles of Ag = 2.96 x 10⁻³ mol
concentration of HBr (Molarity) = conc. of Br (strong acid) =
4.) True:
Since
P(E) = 1 - P(Ec)
P(E)+P(Ec)=1
If you think of it as the chances of you getting e or not getting e, then it's a 100% chance of either one happening.