D.)14 to 16 days ... at least for a fly with is an insect also
plz mark branliest
Answer:
1837.89 Lt
Explanation:
The chemical reaction for this situation is:
NaHCO₃ + HCl → NaCL + H₂O + CO₂ ₍g₎
Where the mola mass we need are:
M NaHCO₃ = 84 g/mol
M CO₂ = 44 g/mol
As we have 6.00 Kg of sodium bicarbonate, then:
6 Kg NaHCO₃ = 71.43 moles of NaHCO₃
Due the stoichiometry of this chemaicl reaction:
1 mol NaHCO₃ = 1 mol CO₂
71.43 moles NaHCO₃ = 71.43 moles CO₂
And considering that CO₂ is an ideal gas, we can use the following formula:
PV=nRT
V = (nRT)/P
n = 71.43 mol
R = 0.083 Ltxatm(molxK)
T = 37°C = 310 K
P = 1 atm
So: V = (71.43x0.083x310)/1
V CO₂ = 1837.89 Lt
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
1st qn - B
2nd qn - A
Last qn - Yes
