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3 years ago

Does anyone know o I'm correct? If not what's the answer?

Chemistry

1 answer:

igomit [66]3 years ago

6 0

I believe it means idea because a Web is out line to a a more structured idea

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Calculate the volume of a 0.323-mol sample of a gas at 265 K and<br> 0.900 atm.

Katarina [22]

Answer:

V = 7.8 L

Explanation:

Message

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3 years ago

What is the pH of 15 mL of .2 M HCl

fenix001 [56]

Easy peasy! All we need to do is plug this formula into our calculator:

-log(M)

So, we'd plug in -log(.2), which is 0.7 :)

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3 years ago

9. Ibuprofen contain which of the following two functional groups: (1 point)

stira [4]

Answer:

A and C

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3 years ago

Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass

Ipatiy [6.2K]

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

= 1.8g/ml

4 0

3 years ago

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0

2 years ago