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:
Hello there!
In this case, according to the given information and chemical equation, it turns out possible for us to calculate the moles of C2O4^2- by firstly setting up the equilibrium expression:
However, according to the question, we just need to apply the given 1:3 mole ratio in the chemical reaction, of iron (III) ions to oxalate ions to obtain:
Regards!
Answer:
molecular eq.
Cs2SO4 (aq) + BaBr2(aq) → BaSO4(s) + 2CsBr(aq)
ionic eq.
SO₄²-(aq) + Ba^+2 (aq) → BaSO4 (s)
explanation
separate the ions
Cs₂^+1(aq) + SO₄^+2(aq) + Ba^+2(aq) + Br₂^-1(aq) → BaSO₄(s) + 2Cs(aq) + Br(aq)
than simplify it
SO₄²-(aq) + Ba^+2 (aq) → BaSO4 (s)
1) (1.00mol H2O2)(34.02g/mol H2O2) = 34.02g H2O2
2) (1.00mol NaNO3)(84.99g/mol NaNO3) = 84.99g NaNO3
3) (8.00mol Ar)(6.02x10²³) = 4.82x10²⁴ atoms
4) (132g Kr) / (83.79g/mol Kr) = 1.58mol Kr
5) (96.4g Si)/(28.09 g/mol Si) = 3.43mol Si
(3.43mol Si)(6.02x10²³) = 2.07x10²⁴ atoms Si
6) You did this one correctly :)
179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms / 1 mol ) = 5.61 x 10^23 atoms of iridium