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3 years ago

I REALLY NEED HELP, PLEASE!!! DUE TONIGHT

Chemistry

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer :

Part 1: The molar mass of $CuCO_3.Cu(OH)_2$ is 222 g/mol

Part 2: The mass of $H_2O$ and $CO_2$ is 18 g and 44 g respectively.

Explanation :

Part 1:

As we know that:

Atomic mass of copper, Cu = 64 g/mol

Atomic mass of carbon, C = 12 g/mol

Atomic mass of oxygen, O = 16 g/mol

Atomic mass of hydrogen, H = 1 g/mol

And we know that:

Molar mass of $CuCO_3.Cu(OH)_2$ = (2 × Atomic mass of Cu) + Atomic mass of C + (5 × Atomic mass of O) + (2 × Atomic mass of H)

Molar mass of $CuCO_3.Cu(OH)_2$ = (2 × 64) + 12 + (5 × 16) + (2 × 1)

Molar mass of $CuCO_3.Cu(OH)_2$ = 128 + 12 + 80 + 2

Molar mass of $CuCO_3.Cu(OH)_2$ = 222 g/mol

Part 2:

The balanced chemical reaction is:

$CuCO_3.Cu(OH)_2(s)\rightarrow 2CuO(s)+H_2O(g)+CO_2(g)$

From the balanced chemical reaction we conclude that:

1 mole of $CuCO_3.Cu(OH)_2$ reacts to give 1 mole of $H_2O$ and 1 mole of $CO_2$

Now we have to determine the mass of $H_2O$ and $CO_2$ .

Mass of $H_2O$ = Moles of

Molar mas of $H_2O$ = 18 /mol

Mass of $H_2O$ = 1 mol × 18 g/mol = 18 g

And,

Mass of $CO_2$ = Moles of

Molar mas of $CO_2$ = 44 /mol

Mass of $CO_2$ = 1 mol × 44 g/mol = 44 g

Therefore, the mass of $H_2O$ and $CO_2$ is 18 g and 44 g respectively.

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Answer: The minimum value of the potential difference through which these ions must be accelerated is $1.87 \times 10^{-5}$ MV.

Explanation:

The given data is as follows.

$m_{1}$ = 35 amu

= $35 \times 1.66 \times 10^{-27}$

= kg

= 37 amu

= $37 \times 1.66 \times 10^{-27}$

= kg

It is known that,

work done on charged particle = gain in kinetic energy

So,

v = $\sqrt{(2 \times q \times \Delta_{V/m})}$

A centripetal force is experienced when charged particles enter a uniform magnetic field.

Hence, F = $q \times v \times B (sin(90))$

$\frac{m \times \frac{v^{2}}{r} = q \times v \times B$

r = $\frac{m \times v}{B \times q}$

And,

$r_{2} = \frac{m_{2} \times v_{2}}{(B \times q)}$

On completion of semi circle, the distance between two paths =

$0.7 \times 10^{-2} = (\sqrt(2 \times m_{2} \times delta_{V/q}) -\sqrt{(2 \times m_{1} \times \frac{\Delta_{V/q}}{B}}$

$0.7 \times 10^{-2} = \sqrt{(\Delta_{V})} \times (\sqrt(2 \times \frac{m_{2}}{q}) - \sqrt{\frac{(2 \times \frac{m_{1}}{q}))}{B}}$

$\sqrt{(\Delta_V)} = 0.7 \times 10^{-2} \times \frac{B}{(\sqrt{(2 \times \frac{m_{2}{q})} - \sqrt{(\frac{2 \times m_{1}}{q}))}$

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$\Delta_V$ = $\sqrt{350}$

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= $1.87 \times 10^{-5}$ MV

Thus, we can conclude that the minimum value of the potential difference through which these ions must be accelerated is $1.87 \times 10^{-5}$ MV.

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