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ahrayia [7]
3 years ago
12

It contains only one type of molecule or extended structure.

Chemistry
1 answer:
Gelneren [198K]3 years ago
6 0
What’s the question though
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If you have 9900 g of donuts, how many donuts do you have if each donut has a mass of 150 g?
AlexFokin [52]

You would have 66 donuts.


If each you have 9900g of donuts and each donut has a mass of 150g, we can divide 9900 by 150 to figure out how many donuts we have. To do this, we first have to figure out how many times 150 goes into 990. It goes into 6 times and has a remainder of 90. Now we have to figure out how many times 150 goes into 900. It goes in 6 times, meaning that our answer is 66 donuts.

8 0
3 years ago
Which one of these elements was suggested as transition element by Mendeleev?
BARSIC [14]

Answer:

The answer is I think None.

5 0
3 years ago
Color is not very helpful in mineral itdentification because
matrenka [14]
Some colours appear as white
7 0
3 years ago
A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol
katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is -17.6^oC

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

6 0
2 years ago
Based on your observations from the glowing splint test with oxygen gas, briefly explain why liquid oxygen is a very hazardous s
Cerrena [4.2K]

Answer:

liquid oxygen is highly flammable

Explanation:

near any source of heat. liquid oxygen can explode to flames thus being hazardous

3 0
3 years ago
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