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weeeeeb [17]
3 years ago
13

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chl

oride, resulting in the precipi- tation of all the sulfate ions as calcium sulfate. the pre- cipitate was collected, dried, and found to weigh 1.36 g. determine the atomic mass of m and identify m.
Chemistry
1 answer:
solong [7]3 years ago
8 0

The reaction between m_{2}SO_{4} with calcium chloride can be shown as-m_{2}SO_{4}+CaCl_{2}→CaSO_{4}↓+2mCl. The molecular weight of CaSO_{4} is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of m_{2}SO_{4} = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 molesm_{2}SO_{4} to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989×10^{-3} moles of calcium sulfate.

Thus, 9.989×10^{-3} moles of m_{2}SO_{4} reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of m_{2}SO_{4} is 2x+96.

So we may write 9.989×10^{-3}× (2x+96) =1.42

Or, 2x + 96 = 142.146

Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).  

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Which reaction type is really part of another type of reaction but treated separately? Explain
Novosadov [1.4K]

Answer:

Oxidation - reduction reaction which is called redox reaction.

Explanation:

The type of reaction that is a part of another type but treated separately is called Oxidation - reduction reaction.

Thus is because they usually involve 2 separate half reactions which are oxidation reactions that involve loss of electrons and then reduction equations which involve gain of electrons.

They are treated separately for example when sodium reacts with chlorine to form sodium chloride.

2Na + Cl2 = 2NaCl

Now, sodium undergoes oxidation by loosing elctrons and it's half reaction is;

2Na → 2Na​^(+) + 2e^(-)

​​Meanwhile chlorine undergoes reduction by gaining electrons and its half reaction is;

Cl2 + 2e^(-) → 2Cl^(-)

6 0
3 years ago
If all three metals were exposed to extremely high temperatures, which would be the last to melt?
Verdich [7]
Copper would be the last to melt because it has the highest melting point!
3 0
2 years ago
Read 2 more answers
At 25◦C a 4 L sample of H2 exerts a pressure of 5 atm. What pressure would the same sample exert in a 2 L container at 25◦C?
kykrilka [37]

Answer:

10 atm.

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 5 atm

P2 = ?

V1 = 4L

V2 = 2L

T1 = 25°C = 25 + 273 = 298K

T2 = 25°C = 298K

Using P1V1/T1 = P2V2/T2

5 × 4/298 = P2 × 2/298

20/298 = 2P2/298

Cross multiply

298 × 20 = 298 × 2P2

5960 = 596P2

P2 = 5960 ÷ 596

P2 = 10 atm.

4 0
3 years ago
g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of
sattari [20]

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

<em>Moles E-stilbene:</em>

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

<em>Mass Product = Theoretical yield:</em>

0.00277 moles * (340.058g/mol) = 0.9433g

4 0
3 years ago
Helium and nickel are examples of ________. These
emmasim [6.3K]

Answer:

Elements, in turn, are pure substances—such as nickel, hydrogen, and helium—that make up all kinds of matter.

Hope This Helps!       Have A Nice Day!!

3 0
3 years ago
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