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Yuri [45]
3 years ago
5

How many moles are in an 8.40 grams of a non electrolyte solution

Chemistry
1 answer:
Crazy boy [7]3 years ago
6 0

Answer: you need to define what is the substance

Which mass is 8.40 g. If you mean you have pure

Water, amount of substance n = m/M = 8.40 / 18.016 g/ mol

= 0.466 mol

Explanation:

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What type of energy does a ball have when it rolls down a hill?
spin [16.1K]
A. Kinetic energy
it is moving
5 0
2 years ago
Complete the equation for the dissociation of Na2CO3(aq).Omit water from the equation because it is understood to be present.
ozzi

Answer:

Na2CO3 <==> 2Na^+ + CO3^2-

Explanation:

6 0
2 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
In one of his experiments, Lavoisier placed 10.0 grams of mercury (II) oxide into a sealed container and heated it. The mercury
inysia [295]

Oxygen gas produced : 0.7 g

<h3>Further explanation</h3>

Given

10.0 grams HgO

9.3 grams Hg

Required

Oxygen gas produced

Solution

Reaction⇒Decomposition

2HgO(s)⇒2Hg(l)+O₂(g)

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

mass of reactants = mass of products

mass  HgO = mass Hg + mass O₂

10 g = 9.3 g + mass O₂

mass O₂ = 0.7 g

4 0
3 years ago
12. How much mass is in a 3.25-mole sample of NH 4 OH? A. 10.8 g B. 34.0 g C. 35.1 g D. 114 g
Galina-37 [17]

Answer:

D. 114 g

Explanation:

  • NH4OH molecular weight. Molar mass of NH4OH = 35.0458 g/mol This compound is also known as Ammonium Hydroxide.
  • Convert grams NH4OH to moles or moles NH4OH to grams. Molecular weight calculation: 14.0067 + 1.00794*4 + 15.9994 + 1.00794.
8 0
3 years ago
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