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2 years ago

Which elements can have expanded octets?

Chemistry

1 answer:

PSYCHO15rus [73]2 years ago

7 0

Answer:

Sulfur, phosphorus, silicon, and chlorine are common examples of elements that form an expanded octet.

Explanation:

Phosphorus pentachloride (PCl5) and sulfur hexafluoride (SF6) are examples of molecules that deviate from the octet rule by having more than 8 electrons around the central atom

You might be interested in

Which statement compares the masses of two subatomic particles

Lelechka [254]

I don't see any options to choose from, but, I have seen this question.

When comparing the masses of two subatomic particles, the mass of a proton is greater than the mass of an electron.

Brainlest answer is always appreciated ( ͡ ͜ʖ ͡ )

4 0

3 years ago

The molarity of a solution containing 7.1 g of sodium sulfate in 100 mL of an aqueous solution is

Sergio [31]

Answer:

0.5 M

Explanation:

First we have to start with the molarity equation:

$M=\frac{mol}{L}$

We need to know the amount of moles and the litters.

If we have 100 mL we can convert this value to “L”, so:

$100~mL\frac{1~L}{1000~mL}$

$0.1~ L$

Now we can continue with the moles, for this we have to know the formula of sodium sulfate $Na_2SO_4$ , with this formula we can calculate the molar mass if we know the atomic mass of each atom on the formula (Na: 23 g/mol, S: 32 g/mol, O: 16 g/mol). We have to multiply each atomic mass by the amount of atoms in the formula, so:

$molar~ mass~=~ (23*2)+(32*1)+(16*4)= ~ 142~ g/mol$

In other words:

$1~mol~ Na_2SO_4=~142~g~ of~Na_2SO_4$

Now we can calculate the moles:

$7.1~g~ of~Na_2SO_4\frac{1~mol~ Na_2SO_4}{142~g~ of~Na_2SO_4}$

$0.05~mol~ Na_2SO_4$

Finally, we can calculate the molarity:

$M=\frac{0.05~mol~ Na_2SO_4 }{0.1~ L}$

$M=0.5$

I hope it helps!

4 0

3 years ago

Read 2 more answers

Calculate how much 95% ethyl alcohol will be required to dissolve 0.1 g of sulfanilamide at 78 c using volume calculate how much

Oduvanchick [21]

According to the source below, the solubility of sulfanilamide in 95% ethyl alcohol at 78°C is 210 mg/mL. Since 0.1 g = 100 mg, we can set up a proportion:

(210 mg) / (1 mL) = (100 mg) / (x mL) Solving, x = 0.48 mL of 95% ethyl alcohol will be required.

I do not know previously the solubility of sulfanilamide in 95% ethyl alcohol. Let us accept the solubility you quoted here.

100/210 = 0.47619047619.. ≈ 0.48 (ml)

at 0C, the amount of sulfanilamide remains in the solution is: 14*(100/210) = 6.67 (mg), since you only have 0.48 ml solution.

The volume of the solution will change a little by cooling from 78C to 0C. You may also consider this volume change if you have data.

6 0

3 years ago

8. The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to:

dedylja [7]

The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to "add water, stir, filter and evaporate the filtrate".

Option: C

Explanation:

The salt and sand can be separated on the basis of solubility, as we know the salt is chemically known as sodium chloride, which have good solubility in water. The most general method is the mixture is taken into a flask or beaker and water is added with stirring, where salt get dissolved and sand remain as it is, because NaCl is capable enough to form hydrogen bonding with water, while sand have absence of such property. Then this solution containing insoluble sand is filtered by using filter paper. The sand is received in filter paper while filtrate in beaker is evaporated by boiling it in order to receive salt as residue.

8 0

3 years ago

What is the density of an object if its mass is 10 g and a volume of 2 mL?

Lilit [14]

Hi there! Let's solve this problem shall we!

⠀Volume = 10g

Mass = 2 mL

In this specific problem, they are asking us to find the density of the object. So, using the information given to us (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

M = 2mL

V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

D = 5 g/mL

*Make sure you put the units for your final solution!*

4 0

1 year ago