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Andrei [34K]
2 years ago
9

5. Which mineral listed below is commonly known as "fools gold"?

Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
4 0

Answer:

b. pyrite

Explanation:

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Based on your graph in the Periodic Trends Lab, what is the relationship between atomic radius and ionization energy ?
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The element born, b, has an atomic mass of 10.81 amu according to the periodic table. however, no single atom of boron has amass
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Read 2 more answers
The freezing point of a nonelectrolyte solution containing 30.0 g of a solute dissolved in 250.0 g of water is observed to be -2
scZoUnD [109]

Answer:

Molar mass of solute is 89.28 g/m

Explanation:

Colligative property of freezing point depression to solve this:

ΔT = Kf . m . i

i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.

m = molalilty (mol of solute/1kg of solvent

ΔT = T° freeze pure solvent - T° freeze solution

0°C - (-2.50°C) = 1.86 °C/m . m

2.50°C / 1.86 m/°C = m

1.34 mol solute/kg solvent = m

This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.

1000 g ____ 1.34 moles

250 g _____(2.50 . 1.34) / 1000 = 0.336 moles

To find the molar mass, we divide mass / moles

30 g/ 0.336 moles = 89.28 g/m

7 0
3 years ago
A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a
Molodets [167]

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of CH_3C0_2 H= 77 \ g \\\\

Molar mass of CH_3C0_2 H = 60.05 \  \frac{g}{mol} \\\\

No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

                                          = 1.28 \ mol \ CH_3CO_2H

Mass of water (H_2O)= 42 \ g

The molar mass of water (H_2O)= 18.02 \  \frac{g}{mol}

moles of water n_{H_2O}:

= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

Molfraction of acetic acid:

X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

                     =\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35

7 0
3 years ago
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