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pochemuha
3 years ago
5

Balance the following equations:C(s) +CO2(g)- CO(g)

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

Balanced equation is

2C + 2C02→ 4CO

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What is the mass of and object with a volume of 100 cm3 and a density of 10 g/cm3?
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Answer:

1000 g

Explanation:

d = m/v

We are given d: 10g/cm3

and v: 100cm3

Plug them into the equation to get 10 = m/100

Then, cross multiply 10x100 to get mass which is: 1000g

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Which best describes the particles in a gas when the temperature rises from 23 °C to 46 °C?
mariarad [96]
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5 0
3 years ago
ASAP!
Andrei [34K]

Answer:

8.3334%

Explanation:

You have two masses. To find the percent of sodium chloride in water by mass, you divide the mass of NaCl by water. First, make both units the same. Easiest is to convert kg into g. 1.5kg = 1500g

125g NaCl/1500g H2O = 0.0833333333 ==> 8.3334%

5 0
3 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
3 years ago
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