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expeople1 [14]
3 years ago
13

Calculate the mass (in grams) of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard, assumi

ng the standard is known to have a hardness of 75.0 ppm (hardness due to CaCO3).
Chemistry
1 answer:
Natali5045456 [20]3 years ago
8 0
In one gram of standard, there are 75 × 10⁻⁶ grams of CaCO3
Thus, in 50 ml, there are 50 grams of standard (assuming density of standard equal to that of water) so grams of CaCO3 are:
50 × 75 × 10⁻⁶
= 3.75 × 10⁻³ grams
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Answer:

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Explanation:

The reaction expression is given as:

          Li   +   F₂    →     LiF

We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.

 Let use a mathematical approach to solve this problem;

  Assign variables a,b and c as the coefficients that will balance the expression:

             aLi   +  bF₂    →     cLiF

Conserving Li: a  = c

                   F:  2b  = c

          let a = 1, c  = 1 and b  = \frac{1}{2}  

  Multiply through by 2;

        a  = 2, b = 1 and c  = 2

               2Li   +   F₂    →    2LiF

 

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3 years ago
a sample of unknown material weighs 500 n in air and 200 n when immesersed in alcholol with a specfic gravity of 0.7 what is the
svet-max [94.6K]

Answer: The mass density is 1166.36 kg/m^{3}.

Explanation:

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Weight of sample in alcohol (F_{alc}) = 200 N

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Formula used to calculate Buoyant force is as follows.

F_{B} = F_{air} - F_{alc}\\= 500 - 200 \\= 300 N

Hence, volume of the material is calculated as follows.

V = \frac{F_{B}}{\rho \times g}

where,

F_{B} = Buoyant force

\rho = specific gravity

g = acceleration due to gravity = 9.81

Substitute the values into above formula.

V = \frac{F_{B}}{\rho \times g}\\= \frac{300}{700 \times 9.81}\\= \frac{300}{6867}\\= 0.0437 m^{3}

Now, mass of the material is calculated as follows.

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Therefore, density of the material or mass density is as follows.

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Thus, we can conclude that the mass density is 1166.36 kg/m^{3}.

7 0
2 years ago
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