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nevsk [136]
3 years ago
15

neon has an average atomic mass of 20.2 g/mol, whereas argon has an average atomic mass of 40.0 g/mol. How would the number of a

toms in a 1.0 mol sample of neon compare to the number of atoms in a 4.0 mol sample of argon
Chemistry
2 answers:
topjm [15]3 years ago
8 0
Argon sample would have 20.2 divided by 40.0
Greeley [361]3 years ago
6 0

Answer : The number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

Explanation :

As we know that the 1 mole of substance contains 6.022\times 10^{23} number of atoms.

As per question,

The 1 mole sample of neon contains 6.022\times 10^{23} number of atoms.

And,

As, 1 mole sample of argon contains 6.022\times 10^{23} number of atoms

So, 4.0 mole sample of argon contains 4.0\times 6.022\times 10^{23}=24.088\times 10^{23} number of atoms

From this we conclude that, the number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

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Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

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We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

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