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Vaselesa [24]
2 years ago
12

Calculate the molar mass for each of the following compounds 7. PbSO

Chemistry
1 answer:
Simora [160]2 years ago
5 0

Answer:

Fe3(PO4)2

Explanation:

calculate the molar mass for each of the following compounds 7. PbSO

8. Ca(OH)2

9. Na3PO4

10. (NH4)2CO3

11. C6H12O6

12. Fe3(PO4)2

13. (NH4)2S

14. Zn(C2H3O2)2

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What is the mass of sodium (Na) in 50 grams of table salt (NaCl)? Show your work.
Margaret [11]

Answer:

19 g

Explanation:

Data Given:

Sodium Chloride (table salt) = 50 g

Amount of sodium (Na) = ?

Solution:

Molecular weight calculation:

NaCl = 23 + 35.5

NaCl = 58.5 g/mol

Mass contributed by Sodium = 23 g

calculate the mole percent composition of sodium (Na) in sodium Chloride.

Since the percentage of compound is 100

So,

Percent of sodium (Na) = 23 / 58.5 x 100

Percent of sodium (Na) = 39.3 %

It means that for ever gram of sodium chloride there is 0.393 g of Na is present.

So,

for the 50 grams of table salt (NaCl) the mass of Na will be

mass of sodium (Na) = 0.393 x 50 g

mass of sodium (Na) = 19 g

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3 years ago
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There are seven petals on that flower.<br> O<br> qualitative<br> quantitative
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3 years ago
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When an atom is in the ground state what must happen for the atom to be in an excited state? what must happen for this atom to r
scoray [572]
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In order for an electron to jump into a higher energy state, it must first absorb energy (heat, light, etc).

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4 0
3 years ago
Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M A
olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are in equilibrium</em>

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are the actual concentrations</em>

<em />

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

<em>Where X is defined as the reaction coordinate</em>

<em />

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

<h3>[Ag⁺] = 2.8x10⁻⁵M</h3>

[CrO₄²⁻] = 0.135M

6 0
3 years ago
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