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andrezito [222]
1 year ago
15

Give the name and symbol or formula of a Group 5A(15) element or compound that fits each description or use:(c) Oxide used as a

laboratory drying agent
Chemistry
1 answer:
Sonja [21]1 year ago
4 0

It is possible to dehydrate neutral and basic gases, amines, low-boiling alcohols, and ethers using calcium oxide, a basic drying agent. its chemical formula is CaO.

CaO cannot be substituted with sulfuric acid. Because the salt ammonium sulfate is created when sulphuric acid and ammonia combine, this is the case. Calcium oxide is a solid that is odorless, white, gray-white, and appears as hard lumps. powerful skin, eye, and mucous membrane irritant. used in fertilizers and pesticides. The group of calcium oxides with a 1:1 ratio of calcium to oxygen includes calcium oxide. It functions as fertilizer.

Learn more about calcium oxide here-

brainly.com/question/14259922

#SPJ4

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
2 years ago
Helppp mee plz guysss
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Answer:

11 and a half

Explanation:

well it luks to be

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yKpoI14uk [10]

Answer:

Balancing the equation

2KMnO₂+10KCl+8H₂SO₄⇒2MnSO₄+6K₂SO₄+8H₂O+5Cl₂

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What are the three challenges of a cell?
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<span>The cell must exchange materials with the environment across the surface membrane. An increase in size will result in a relatively greater increase in volume and mass than in surface area, so that the cell will lose effective exchange capacity. 

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Draw the structure 2 butylbutane
k0ka [10]

Answer:

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Explanation:

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