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avanturin [10]
3 years ago
15

A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac

ed in a second volume of water. bcould not be decomposed into any simpler substances. evaporation of the water from the soluble portion of ayielded 6 grams of solid,
c. heating bin air yielded 8 grams of gas, d, leaving no solid residue. melting and electrolysis of cyielded 2.4 grams of metal, e, and 3.6 grams of gas, f. classify each material described above by placing the corresponding letter in the appropriate place in the table below.
Chemistry
1 answer:
kicyunya [14]3 years ago
3 0
Sksksk isjsjs jsjsjs jsjsjs jsjsjs jsjsjs ksjss
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Why the is ratio of magnesium ions to chloride ions in MgCl2 is 1:2?
Monica [59]
You should read up on Proust's law, better known as the Law of Definite Proportions. This is a chemical law that defines your question more generally, on why the ratio of elements and ions are always fixed.

Basically, this compound Magnesium(II) Chloride is MgCl2 because it has the same number of protons, neutrons, and electrons all the way. This defines the properties of the compound or atom. 
5 0
3 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
3 years ago
How many moles of hydrogen are needed to produce 13.78 mol of ethane?
mamaluj [8]
Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
4 0
3 years ago
Read 2 more answers
How mant grams are there in 7.15 x 10^24 molecules of water?
worty [1.4K]

Answer:

B. 214.02

Explanation:

1 mol of water weighs 18.015 gm and contains 6.023 × 10²³ molecules

From question, We have 7.15 × 10²⁴ molecules

Dividing we get (7.15 × 10 ²⁴) ÷ ( 6.023 × 10²³) = 11.871 molecules

Now, Weight of water = 11.871 × 18.015 = 213.85 which is nearer to option B

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