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:
Explanation:
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
PV =nRT
at STP P= 1atm. T= 273 K
n is the number of moles. O2 has a molar mass of 32.
30 gm of O2 is 30/32= 0.94 =n
PV = nRT
at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K
SO
1 X V = 0.94 X 0.082 X 273
using high school freshman algebra,
V= 0.94 X 0.082 X 273 = 21L
using high school algebra I,
V=
Answer:
1.54×10^20 atoms
Explanation:
It's done by using the relation
N=n×L where,
N = number of entities present
n= amount of substance (mole)
L= Avogadro's constant which is 6.02×10^23
now from the question, given
n=0.000256
And L=6.02×10^23
N= 0.000256×6.02×10^23
N= 1.54×10^20 atoms
Answer:
2S +3O2 =2SO3
Explanation:
2 at the front of sulphur is to equalize the 2 put in SO3.
The first half reaction is:
Sn+ -> Sn2+
You add 1 electron on the right side:
Sn+ -> Sn2+ + 1e-
The second half reacation is:
Ag+ -> Ag
You add 1 electron to the left side:
Ag+ + 1e- -> Ag
Then, you combine both reactions to get the balanced overall reaction since both half reactions have 1 electrons on the right and left side, respectively. You get:
Sn+ + Ag+ -> Sn2+ + Ag
