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Sphinxa [80]
3 years ago
11

Which of the following statements describes the correct method of preparation of 1.00 L of a 2.0 M urea solution?

Chemistry
1 answer:
Zolol [24]3 years ago
8 0

Answer:

To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

Explanation:

Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.

So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles

We know, number of moles of a compound is the ratio of mass to molar mass of that compound.

So, mass of  2 moles of urea = (2\times 60.06)g=120 g

Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

So, option (C) is correct.

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Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block.
SSSSS [86.1K]

The kinetic energy formula is;

  • KE=\frac{m.v^2}{2}

The variable m represents the mass. Its unit is kilogram. We are informed that the mass of the object is 15kg. The variable v represents the linear velocity. Its unit is meter per second. We are informed that the linear velocity of the object is 20m/s. Let's find the kinetic energy of the object by substituting these values in the formula.

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Vanyuwa [196]

Answer: Hello the compound is missing but I was able to get the Full question and missing compound . ( compound = copper sulfate )

<em>answer</em> : statement ; 2 , 3 and 5

Explanation:

The true statements regarding the coordination compound ( copper sulfate ) are :

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During the coordination of compounds dative bonds exits between the transition metals and the Ligands molecules

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We heavier the object the less friction that exist is this true or false
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What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol
statuscvo [17]

Answer:

58.92 g EDTA

Explanation:

315.1 mL = .3151 L    

M = Moles / Liter

.3151 L  x <u>0.5 mol EDTA</u>  x  <u>374 g EDTA</u>  =  58.92 g EDTA

                1 L EDTA             1 mol EDTA

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