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kkurt [141]
3 years ago
11

Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3

7.5 g 100.0 g
Chemistry
2 answers:
Vadim26 [7]3 years ago
7 0
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
Maslowich3 years ago
7 0
The correct answer is letter c 


hope i helped
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A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the
Kamila [148]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

 <u>   P ∝ T</u>

(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

   T_{2} =\frac{333*4}{1.8}

    T_{2} = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0
3 years ago
Part C<br> Number of molecules in 8.437x10-2 mol C6H6
N76 [4]

Answer:

There are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

Explanation:

In this problem, we need to find the number of molecules in 8.437\times 10^{-2} mol of C_6H_6.

The molar mass of C_6H_6 is 6\times 12+1\times 6=78\ g/mol

No of moles = mass/molar mass

We can find mass from above formula.

m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g

Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

Hence, there are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

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3 years ago
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VashaNatasha [74]

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

         H

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H₂N - C - C =O

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The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

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kogti [31]

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