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Oksi-84 [34.3K]
3 years ago
7

Crystals of sodium chloride were prepared by the following method.

Chemistry
1 answer:
Kryger [21]3 years ago
8 0

Answer: the answer is C

IF U MIGHT NEED AN EXPLANATION FEEL FREE TO COMMENT

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Each diagram below represents the nucleus of an atom. How many different elements are represented by the diagrams
Korolek [52]
If the question is multiple choice the answer would be:
2) 2
5 0
3 years ago
A 4 kg rock is rolling 10 m/s. Find its kinetic energy
Pavel [41]
KE = 1/2mv^2
KE = 1/2(4)*10 m/s^2
KE = 2 * 10
KE = 20 joules

Answer: 20 joules 
8 0
3 years ago
What product is formed when benzene reacts with isobutyl chloride in the presence of AlCl3?
oee [108]

The product formed when benzene reacts with isobutyl chloride in the presence of AlCl3 is tertiary butyl benzene.

THE CHEMICAL REACTION AS

step 1 : CH₃ - CH - CH₂Cl +  AlCl₃  →  CH₃ - CH₂ -CH₂ - CH₂⁺ + AlCl⁻

step 2: CH₃ - CH₂ - CH₂ - CH₂⁺   ---H SHIFT--→  (CH₃)₃C⁺

PRODUCT

C₆H₆  + (CH₃)₃ C⁺  → C₆H₆ ( CH₃)₃ C⁺

Hence the product formed is tertiary butyl benzene.

Learn more about chemical reaction on

brainly.com/question/11231920

#SPJ4

4 0
2 years ago
What is the molarity (M) of an 85.0-mL solution (total volume) containing 1.77 g of ethanol (C2H5OH)
Liono4ka [1.6K]

Answer: The molarity of solution is 0.453 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of C_2H_5OH = \frac{\text {given mass}}{\text {Molar mass}}=\frac{1.77g}{46g/mol}=0.0385mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.0385\times 1000}{85.0}

Molarity=0.453

Therefore, the molarity of solution is 0.453 M

7 0
3 years ago
phosporus can be prepared from calcium phosphate by the following reaction: 2Ca3(PO4)2+6SiO2+10C → 6CaSiO3+P4+10CO Phosphorite i
morpeh [17]

Answer:

285 g of P₄

Explanation:

Let's consider the following balanced equation.

2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + P₄ + 10 CO

We know the following relations:

  • 100 g of phosphorite contain 75 g of Ca₃(PO₄)₂
  • 2 moles of Ca₃(PO₄)₂ produce 1 mole of P₄
  • The molar mass of Ca₃(PO₄)₂ is 310 g/mol
  • The molar mass of P₄ is 124 g/mol

Then, for 1.9 kg of phosphorite:

1900g(phosphorite).\frac{75gCa_{3}(PO_{4})_{2}}{100g(phosphorite)} .\frac{1molCa_{3}(PO_{4})_{2}}{310gCa_{3}(PO_{4})_{2}} .\frac{1molP_{4}}{2molCa_{3}(PO_{4})_{2}} .\frac{124gP_{4}}{1molP_{4}} =285gP_{4}

6 0
3 years ago
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