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2 years ago

The fifth element in group 7 in astatine would you expect it to be gas, liquid, or solid? why?

Chemistry

1 answer:

Ymorist [56]2 years ago

6 0

Answer:

solid

Explanation:

Melting and boiling points of Group 7 elements State at room temperature Room temperature is usually taken as being 25°C. At this temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine and astatine are solids. There is therefore a trend in state from gas to liquid to solid as you go down the group.

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andrey2020 [161]

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3 years ago

Which of the following is not a characteristic of a covalent compound?

TEA [102]

'It is formed when metal atoms lose electrons to nonmetal atoms' is the incorrect statement. This statement is the definition of ionic bonding.

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3 years ago

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$

$0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}$

$\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol$

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$ (3)

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$

$\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol$

Solving for $m_{LiOH}$ , we have:

$m_{LiOH} = 0.082 g$

Hence, the percent lithium hydroxide is (eq 1):

$\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%$

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

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I hope it helps you!

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2 years ago

What atoms are present In the molecule KNO3 and how many are there of each ?

nadya68 [22]

Answer:

one potassium, one nitrogen and three oxygen

Explanation:

potassium is K, nitrogen is N and oxygen is O

4 0

3 years ago

A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

Answer: The molar mass of the unknown compound is 223.2 g/mol

Explanation:

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

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

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

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

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

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

Given mass of unknown compound = 15.2 mg = 0.0152 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0

3 years ago