P₁ = 0.90 atm
V₁ = 50.0 mL
T₁ = 298 K
P₂ = 1 atm
T₂ = 273 K
V₂ = P₁ x V₁ x T₂ / T₁ x P₂
V₂ = 0.90 x 50.0 x 273 / 298 x 1
V₂ = 12285 / 298
V₂ = 41 mL
Answer (1)
hope this helps!
Answer:
B. 0.971 g
Explanation:
When MgCl₂(aq) reacts with Pb(NO₃)₂(aq), PbCl₂(s) and Mg(NO₃)₂(aq) are produced:
MgCl₂(aq) + Pb(NO₃)₂(aq) →, PbCl₂(s) + Mg(NO₃)₂(aq)
Thus, we need to find imiting reactant finding moles of each reactant:
<em>Moles MgCl₂:</em>
15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles
<em>Moles Pb(NO₃)₂:</em>
37.5mL = 0.0375L * (0.250mol / L) = 9.38x10⁻³ moles
As the ratio of the reactants is 1:1, the moles of PbCl₂ are 3.48x10⁻³ moles.
We need to convert thes moles to mass using molar mass of PbCl₂ (278.1g/mol), thus:
3.48x10⁻³ moles * (278.1g/mol) =
0.968g of PbCl₂ are precipitate
Thus, right answer is:
<h3>B. 0.971 g</h3>
Answer:
0.32×10⁻³ mol
Explanation:
Given data:
Number of atom of gold(II) chloride = 1.9×10²⁰ atoms
Number of moles = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
1.9×10²⁰ atoms × 1 mol / 6.022 × 10²³ atoms
0.32×10⁻³ mol
You can form 9.20 L of NH₃ gas at 93.0 °C and 37.3 kPa or 2.56 L at STP. The balanced chemical equation is N₂ + 3H₂ → 2NH₃ Since all substances are gases, we can use litres instead of moles in our calculations. You don't specify the pressure and temperature of the NH₃, so I will calculate its volume in two ways.
S, P.
S is a sphere
P is an infinity
