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Bezzdna [24]
2 years ago
6

90 is 50% of what number? Part = Whole x Percent P: W: %

Chemistry
1 answer:
Arlecino [84]2 years ago
4 0

Answer:

90 is 50% of 180

Explanation:

Just multiply 80 by 2 for 50%

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This combination in non polar.
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1. A student collected the following data for a fixed volume of gas: Temperature (⁰C) Pressure (mm of Hg) 10 726 20 750 40 800 7
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Answer is: the missing pressure is 1088.66 mmHg.

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 960 mmHg; pressure of the gas.

T₁ = 100°C + 273.15.

T₁ = 373.15 K; temperature of the gas.

T₂ = 150°C + 273.15.

T₂ = 423.15 K.

p₂ = p₁T₂/T₁.

p₂ = 960 mmHg · 423.15 K / 373.15 K.

p₂ = 1088.66 mmHg.

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Tellurium is a period 5 chalcogen. Selenium is a period 4 chalcogen. If the only factor affecting ionization energies was the nu
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2 years ago
Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is
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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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