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3 years ago

11

Calculate the number of moles of caco3 (calcium carbonate, or limestone) in a 20.0g sample of this substance

1 answer:

Brrunno [24]3 years ago

6 0

First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.

molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃

Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃

<em>Answer: 0.2 moles</em>

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1 year ago

Part d based on your answer to part c, how many bonding electrons and lone pair electrons (nonbonding electrons) are there in th

elena55 [62]

The Lewis Structure of HCN is shown below,

Number of Bonding Electrons:
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Number of Bonding Electrons = 8

Number of Non-Bonding Electrons:
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3 years ago

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3 years ago

Another carbon isotope has six protons and seven neutrons in its nucleus. What do you think this carbon isotope is called?

insens350 [35]

Answer:

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8 0

2 years ago

Read 2 more answers

If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution

Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution = $5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

( $1\mu M=10^{-6} M$ )

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$

8 0

3 years ago