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Kisachek [45]
3 years ago
15

What volume is occupied by 2.33 moles of an ideal gas at 0.953 atm

Chemistry
2 answers:
vfiekz [6]3 years ago
6 0
B is the correct answer I’m pretty
nekit [7.7K]3 years ago
4 0

Answer:

WhAT

Explanation:

WTH

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Read 2 more answers
A sample of the compound magnesium oxide is synthesized as follows. 60. g of magnesium is burned and produces 100. g of magnesiu
Inga [223]

Answer:

30 g of magnesium would be combined with 20 g of oxygen. The law used solving this problem is the Lavoisier Law of conservation of mass.

Explanation:

If 60 g of magnesium combines with 40 g of oxygen to make 100 g of magnesium oxide, then 30 g of magnesium will combine with 20 g of oxygen to make 50 g of magnesium oxide.

This happens because in a chemical reaction there is no creation or descruction of atoms, only a rearrangement. Therefore, the mass of reactants should be equal to the mass of products.

The following equation represents the proportions of the substances:

Mg + 1/2O₂ → MgO

5 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
How many molecules of NH3 are produced from 4.72x10 negative 4 power g of H2?
Delvig [45]
<span>9.40x10^19 molecules.
   The balanced equation for ammonia is:
 N2 + 3H2 ==> 2NH3
   So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
 Atomic weight nitrogen = 14.0067
  Atomic weight hydrogen = 1.00794

   Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol

   Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
   Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol

Now to convert from moles to molecules, just multiply by Avogadro's number: 1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
   Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
7 0
3 years ago
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