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
Presence of single C-C bond
(unsaturated has C=C)
Answer:
C₂ = 0.056 ppm
Explanation:
Given data:
Initial volume = 2.0 mL
Initial concentration = 7.0 ppm
Final volume = 250.0 mL
Final concentration = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Initial concentration
V₁ = Initial volume
C₂ = Final concentration
V₂ = Final volume
Now we will put the values in formula.
C₁V₁ = C₂V₂
7.0 ppm × 2.0 mL = C₂ × 250.0 mL
C₂ = 14.0 ppm.mL /250.0 mL
C₂ = 0.056 ppm
Answer:
protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.
