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3 years ago

What is the name given to the ions of the halogens on the periodic table?

Chemistry

1 answer:

WARRIOR [948]3 years ago

7 0

Answer:

halides

Explanation:

This is one electron away from having a full octet of eight electrons, so these elements tend to form anions having -1 charges, known as halides: fluoride, F-; chloride, Cl-, bromide, Br-, and iodide, I-. In combination with other nonmetals, the halogens form compounds through covalent bonding.

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Zn + 2HCl ---> ZnCl2 + H2 1. Is the equation above balanced or unbalanced? How do you know? (Hint: How many atoms of each typ

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3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

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3 years ago

A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is th

lara [203]

Answer is: the partial pressure of the helium gas is 0.158 atm.

p(mixture) = 0.48 atm; total pressure.

m(H₂) = 1.0 g; mass of hydrogen gas.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 1.0 g ÷ 2 g/mol.

n(H₂) = 0.5 mol; amount of hydrogen.

m(He) = 1.0 g; mass of helium.

n(He) = 1 g ÷ 4 g/mol.

n(He) = 0.25 mol; amount of helium.

χ(H₂) = 0.5 mol ÷ 0.75 mol.

χ(H₂) = 0.67; mole fraction of hydrogen.

χ(He) = 0.25 mol ÷ 0.75 mol.

χ(He) = 0.33; mole fraction of helium.

p(He) = 0.33 · 0.48 atm.

p(He) = 0.158 atm; the partial pressure of the helium gas.

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3 years ago