Ask question

Ask question

All categories

3 years ago

14

Copper2 nitrate > nitrogen dioxide + copper2 oxide+ oxygen

1 answer:

erastovalidia [21]3 years ago

3 0

Cu(NO3)2>NO2+CuO+O2 balanced: 2Cu(NO3)2=4NO2+2CuO+O2

You might be interested in

Identify the name of the smallest particle of an element that still retains all the chemical properties of that element?

Nezavi [6.7K]

The smallest particle of an element that still retains the chemical properties of it is an atom.

4 0

3 years ago

A compound is formed when 12.2 g Mg combines completely with 5.16 g N. What is the percent composition of this compound? Please

STALIN [3.7K]

3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷<span>24,4g/mol=0,5mol-limiting reagent
</span>n(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.

4 0

3 years ago

Which two have to do with Movement of Galaxies? (pick 2)

pychu [463]

I don’t see anything

6 0

3 years ago

Which describes Lithium chloride

Rasek [7]

Answer:

Lithium chloride is a white solid hygroscopic soluble in water.

8 0

3 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?

Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

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

<u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

<u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

3 years ago