Do you think the video game "Minecraft" will ever die?

What is the molar mass of Mg(NO2)2?

qwelly [4]

Answer:

116g/mol

Explanation:

Mg=24

NO2=46

multiple 46 by 2and then add 24

3 0

2 years ago

0.352 g sample of a diprotic acid is dissolved in water and titrated with 0.150 M NaOH.0.150 M NaOH. What is the molar mass of t

Aliun [14]

<h3>Answer:</h3>

128.94 g/mol

<h3>Explanation:</h3>

Given;

Mass of diprotic acid, H₂X =0.352 g

Molarity of NaOH = 0.150 M

Volume of the NaOH = 36.4 mL

We are required to calculate the molar mass of the acid.

Note: A diprotic acid is an acid that contains 2 replaceable hydrogen atoms

<h3>Step 1: Write the balanced equation fro the reaction;</h3>

The balanced equation for the reaction between the diprotic acid and NaOH will be;

H₂X(aq) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l)

<h3>Step 2: Determine the number of moles of NaOH used </h3>

Given the molarity and volume of NaOH we can calculate the number of moles;

Moles of NaOH = Molarity × Volume

= 0.150 M × 0.0364 L

= 0.00546 moles

<h3>Step 3: Use the mole ratio to determine moles of the acid </h3>

From the equation;

1 mole of the acid reacts with 2 moles of NaOH

Therefore; H₂x : NaOH = 1 : 2

Moles of H₂x = 0.00546 moles ÷ 2

= 0.00273 moles

<h3>Step 4: Determine the molar mass of the acid.</h3>

Molar mass is the mass equivalent to 1 mole of a compound or element

From our calculations;

0.00273 moles = 0.352 g of the acid;

Therefore, mass in 1 mole ;

= 0.352 g ÷ 0.00273 moles

= 128.94 g/mol

Thus, the molar mass of the diprotic, H₂X is 128.94 g/mol

7 0

3 years ago

QUESTION 17

borishaifa [10]

Question 17:

False; cutting hair would change what it looks like but, braiding it keeps it the same substance it is.

Question 18 :

My best guess would either be A or D. I would lean more towards D because if there are different mixtures then it depends on what you are mixing.

Hope it helps in some way. <3

5 0

2 years ago

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

<u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

<u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0

3 years ago

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =

Lady bird [3.3K]

Answer:

$4.86\times10^{-7}\ \text{m}$

Explanation:

R = Rydberg constant = $1.09677583\times 10^7\ \text{m}^{-1}$

$n_1$ = Principal quantum number of an energy level = 2

$n_2$ = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

The wavelength of the light emitted is $4.86\times10^{-7}\ \text{m}$ .

6 0

3 years ago