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Nady [450]
2 years ago
9

Convert 18.9 moles to MgCl2 to formula units

Chemistry
1 answer:
Masja [62]2 years ago
3 0

Answer:

18.9 moles of MgCl2 = 17.834 kg of MgCl2

Explanation:

The molecular weight of MgCl is 80.0 g/mol . So, to convert the given mole amount to grams, multiply this by this number, which is constant for all compounds with a specific composition (mass fraction).

Considering the original question was in the context of chemistry, I wanted to make it seem formal and more educational too. Hopefully that worked!  

EDIT: Came up with some text around what happens inside cells that would have made it better if someone just had an issue converting units, but I doubt my answer will be accepted >.<

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At a certain temperature Kc = 9.0 for the equilibrium 24() ⇔ 22(). What is
Iteru [2.4K]

Answer:

.

Explanation:

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5 0
3 years ago
What is the cost of coal in dollars per ton if it costs $0.04 per kilogram?
Anettt [7]

Answer:

36.2874 $

Explanation:

1 ton = 907.185 kilograms.

907.185 x 0.04 =

36.2874

<em>-kiniwih426</em>

8 0
3 years ago
How do you find the atoms of nitrogen in this substance (NH4)2CO3?
Katen [24]
6.022*10^23 is the answer
3 0
3 years ago
A sample of C3H8 has 6.72 x 10^24 H atoms. <br> What is the total mass of the sample?
gregori [183]
The answer is 492.8 g


1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample. 

1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole    

6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n

n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol


2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:

Ar(C) = 12 g/mol

Ar(H) = 1 g/mol

Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol



3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:

m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g

6 0
2 years ago
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
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