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erik [133]
3 years ago
7

How many grams are in a 3.1E21 atom sample of vanadium

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
There are 50.9415 grams in vanadium


hope that helps :)
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A sample of nitrogen gas is collected over water at temperature of 20.0˚C. What is the pressure of the nitrogen gas if atmospher
Levart [38]

Answer:

P_N=0.987atm

Explanation:

Hello there!

In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

P_T=P_w+P_N

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

1.01atm=17.5torr*\frac{1atm}{760torr} +P_N\\\\P_N=1.01atm-0.023atm\\\\P_N=0.987atm

4 0
3 years ago
Explain what is meant by the phrase 'the heat death of the universe."
Vikki [24]

Answer:

See explanation.

Explanation:

Are you literally posting your entire you chemistry homework on this site, one question at a time? Anyways, the heat death refers to the second law of thermodynamics and entropy. Heat is constantly flowing from warmer to cooler objects and never the other way around. This heat flow increases entropy, which is constantly increasing. The universe will eventually disperse all of its heat energy away to continuously increase entropy and reach a limit as the temperature reaches 0 K at which point all molecular motion will cease and so will the life of the universe.

6 0
1 year ago
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An athlete can run 9 kilometers in 1 hour. If the athlete runs at that same average speed for 30 minutes, how far will the athle
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The answer to the problem is 4.5 kilometers. you can solve this problem by cross multiplying
4 0
2 years ago
Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

7 0
2 years ago
The radius of a vanadium atom is 130 pm. How many vanadium atoms would have to be laid side by side to span a distance of 1.30 m
monitta

Answer:

5 000 000 (5 million atoms)

Explanation:

Let us assume that a vanadium atom has a spherical shape.

diameter of a sphere = 2 x radius of the sphere

Thus,

Radius of a vanadium atom = 130 pm

                                              = 130 x 10^{-12} m

The diameter of a vanadium atom = 2 x radius

                                                         = 2 x 130 x 10^{-12}

                                               = 260 x 10^{-12} m

Given a distance of 1.30 mm = 1.30 x 10^{-3} m,

The number of vanadium atoms required to span the distance = \frac{1.3*10^{-3} }{260*10^{-12} }

                                                  = 5000000

Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.

3 0
3 years ago
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