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:
n = 0.594 mol
Explanation:
Parameters of the question.
Heat = 979J
Initial Temperature (T1) = 10.6 C = 283.6 K (converting to Kelvin by adding 273)
FinalTemperature (T2) = 26.4 C = 299.4 K (converting to Kelvin by adding 273)
ΔT = T2 -T1 = 299.4 - 283.6 = 15.8K
Work done (W) = 221 J
number of moles (n) = ?
Formular to be used;
W=PΔV
From ideal gas equation,
PV = nRT
PΔV = nRΔT
where R = gas constant = 8.314 J / mol·K
Sunstituting in the work done equation,
we have W=nRΔT
Making n subject of formular,
![n = \frac{RT}{W}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BRT%7D%7BW%7D)
n = [8.314 * 15.8]/221
n = 131.36/221
n = 0.594 mol
Answer:
Carboxylic acid, any of a class of organic compounds in which a carbon atom is ... to an oxygen atom by a double bond and to a hydroxyl group by a single bond. ... sour-milk products) and citric acid (found in citrus fruits), and many keto acids ...
Explanation:
Answer:
open system
Explanation:
it needs to loose mass not gain mass