All categories

3 years ago

What is the molar mass of 3.45 moles of NO2

Chemistry

2 answers:

marissa [1.9K]3 years ago

4 0

I Believe it would be 46.005500

hoped this helped let me know if I'm wrong

Vinil7 [7]3 years ago

4 0

Answer: The mass of nitrogen dioxide for given number of moles are 158.7 grams.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Moles of nitrogen dioxide = 3.45 moles

Molar mass of nitrogen dioxide = $[14+(2\times 16)]=46g/mol$

Putting values in above equation, we get:

$3.45mol=\frac{\text{Mass of }NO_2}{46g/mol}\\\\\text{Mass of }NO_2=(3.45mol\times 46g/mol)=158.7g$

Hence, the mass of nitrogen dioxide for given number of moles are 158.7 grams.

You might be interested in

Express In scientific notation. Remember, M must be a number 1 SM <10. 18 =

Nesterboy [21]

Answer: 1.8 x 10^1 or the scientific e notation is: 1.8e1

6 0

3 years ago

Read 2 more answers

Nothing sus please haha http://quizprank.xyz/message.php?id=n8q76jgf

SVEN [57.7K]

Answer:

lol thanks for the point btws! i really needed them for my math test questions have a good day btw! good vibes!

Explanation:

8 0

3 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

2 years ago

Balance the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method.

LuckyWell [14K]

First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.

Mg --> Mg+2

Hg+2 --> Hg+1

Then you have to balance each half reaction first with electrons before adding them together in one equation

$Mg$ ⇒ $Mg^{+2} + 2e^{-1}$

and

$2Hg^{+2} + 2e^{-1}$ ⇒ $Hg_{2} ^{+2}$

and then combine them together to form

$Mg + 2 Hg^{+2} + 2e$ ⇒ $Mg^{+2} + 2e ^{-} + 2 Hg^{+1} + 2 e^{-}$

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: Add H+ and H2O to balance the H's and O's in acidic solution if needed.

8 0

3 years ago

Lipid vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high con

NNADVOKAT [17]

Answer:

The question is not complete, the complete question should be "Lipids vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high concentration of solutes, the solution outside the vesicle is said to be Hypertonic. True or False"

The answer is True

Explanation:

This is because it contains greater concentration of solutes on the outside of the cell than the increase.

In other words hypertonic solutions have more concentrate of solutions on the outside than the inside.

3 0

2 years ago