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mario62 [17]
3 years ago
15

Please help Which of the following is the SI unit used to measure length? a.kilogram b.liter c.meter d.Kelvin

Chemistry
1 answer:
Oksanka [162]3 years ago
6 0
<h3><u>Answer</u> :</h3>

A. kilogram is the SI unit used to measure mass of body.

  • <u>1 kilogram = 100 gram</u>

B. litre is the SI unit used to measure volume of substance.

  • <u>1 litre = 10ˉ³ m³</u>

C. meter is the unit used to measure length of body.

  • <u>1 met</u><u>e</u><u>r</u><u> = 100 centimeters</u>

D. Kelvin is the unit used to measure temperature of body.

  • <u>1 K = -273.15 °C</u>

Hence, (B) is the correct answer!

Cheers!

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What is the percent yield if the theoretical yield is 256,5 g and the
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Answer:

78.9% yield

Explanation:

% Yield = Actual Lab Yield/Theoretical Yield x 100%

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3 years ago
Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.52 × 10−4 g MgCl2 in 2.00 L of water
dybincka [34]

Answer:

Explanation:

Given parameters:

Mass of MgCl₂ = 2.52 x 10⁻⁴g

Volume of water = 2L

Unkown:

Concentration of MgCl₂ =?

Concentration of Mg²⁺ = ?

Concentration of Cl⁻ =?

Solution:

  Concentration is defined as the number of moles of a solute contained in a solution.

   Concentration = \frac{number of moles }{volume}

 To find the number of moles"

           number of moles = \frac{mass}{molar mass}

   Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol

       number of moles = \frac{0.000}{molar mass} = 0.000252moles

 Concentration of MgCl₂ = \frac{0.000252}{95.3} = 2.64 x 10⁻⁶moldm⁻³

 

from the formula of the compound;

    1 mole of MgCl₂ contains 1 mole of Mg²⁺

Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg

Also;

   1 mole of MgCl₂ contains 2 mole of Cl⁻

  2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³

 

Expressing in ppm;

               1ppm = 1mg/L

   2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺

2.64 x 10⁻⁶moldm⁻³  = 2.64 x 10⁻⁶moldm⁻³  x molar mass(g/mol)

                                     = 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L

    g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 =   6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm

5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;

   5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L

   g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm

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Calculate the ph of 0,24 m of kch3coo.? ​
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Answer:

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[H

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here is your answer if you like my answer please follow

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