Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
Answer:
Iterate to improve the solution
Explanation:
apex
Answer:
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
Explanation:
The general schemefor a reaction is given as;
Reactants --> Products
In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.
The equation is given as;
AgNO3 + Na2SO4 --> Ag2SO4
The other poduct formed in this reaction is NaNO3.
The full reaction is given as;
AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3
The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.
The balanced equation is given as;
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
In this equation, there are equal number of moles of the atoms on both sides.
5.18mL i hope this helps i hope this does to!
Answer:
The answer to your question is 122.4 g of O₂
Explanation:
Data
mass of O₂ = ?
moles of H₂O = 7.65
Process
1.- Write the balanced chemical reaction
2H₂O ⇒ 2H₂ + O₂
2.- Convert the moles of H₂O to grams
molar mass of H₂O = 2 + 16 = 18 g
18 g of H₂O ---------------- 1 mol
x ----------------- 7.65 moles
x = (7.65 x 18) / 1
x = 137.7 g H₂O
3.- Calculate the grams of O₂
36 g of H₂O -------------------- 32 g of O₂
137.7 g of H₂O ------------------- x
x = (32 x 137.7) / 36
x = 122.4 g of O₂