Hey there!:
<span>Reaction stoichiometry :
</span>
Number of moles NaHCO3 :
Molar mass NaHCO3 = <span>84.007 g/mol
</span>
n = m / mm
n = 2.50 / 84.007
n = 0.0297moles of NaHCO3
2 NaHCO3 + H2SO4 = Na2SO4 + 2 CO2 + 2 H2O
2 moles NaHCO3 ----------------- 1 mole H2SO4
0.0297 moles NaHCO3 ----------- moles H2SO4
moles H2SO4 = 0.0297 * 1 / 2
moles H2SO4 = 0.0297 / 2
= 0.01485 moles of H2SO4
Therefore:
Molarity ( H2SO4 ) = moles H2SO4 / volume
0.600 M = 0.01485 / V
V = 0.01485 / 0.600
V = 0.02475 L of H2SO4
hope this helps!
Answer:
2.83 × 10²² molecules.
Explanation:
Using the formula below to calculate the number of moles of Naphthalene:
mole = mass/molar mass
Where;
mass = 6.00g
Molar mass = 128.18 g/mol
mole = 6/128.18
mole = 0.047mol
To calculate the number of molecules of naphthalene in the mothballs, we multiply the number of moles (n) by Avagadro's number i.e.
number of molecule = n × 6.02 × 10^23
number of molecules = 0.047 × 6.02 × 10^23
= 0.283 × 10²³
= 2.83 × 10²² molecules.
C₆H₁₂O₆ + 6 O₂ ⇒ 6 CO₂ + 6 H₂O
This is an oxidation-reduction reaction, because oxidation numbers of carbon and oxygen atoms are different than before reaction.
(-_-(-_-)-_-)