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3 years ago

Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:

Chemistry

1 answer:

nadezda [96]3 years ago

8 0

Answer : The number of silver atoms produced will be $39.66\times 10^{23}$

Explanation : Given,

Number of atoms of copper = $19.83\times 10^{23}$

The given balanced chemical reaction,

$Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag$

From the balanced chemical reaction we conclude that,

As, $6.02\times 10^{23}$ number of copper atoms react to give $2\times 6.02\times 10^{23}$ number of silver atoms.

So, $19.83\times 10^{23}$ number of copper atoms react to give $2\times \frac{6.02\times 10^{23}}{6.02\times 10^{23}}\times 19.83\times 10^{23}=39.66\times 10^{23}$ number of silver atoms.

Therefore, the number of silver atoms produced will be $39.66\times 10^{23}$

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A tank at is filled with of chlorine pentafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal

Ivan

Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

= 44.53 kPa

Total Pressure exerted by the gases = 60.58 kPa

Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

Mass = 13.3 g

Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

This is Dalton's law of Partial Pressure.

- Mole fraction of Chlorine Pentafluoride

= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

3 0

3 years ago

What is the atomic mass of an atom that has 6 protons, 6 neutrons, and 6 electrons? A) 6 B) 8 C) + 1 D) 12 E) 18

Oksana_A [137]

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Atomic mass of an atom is defined as the sum of number of neutrons and number of protons that are present in an atom. It is represented as 'A'.

Atomic number = Number of protons + Number of neutrons

We are given:

Number of protons = 6

Number of neutrons = 6

Number of electrons = 6

Atomic mass = 6 + 6 = 12

Hence, the correct answer is Option D.

6 0

3 years ago

Please help I will reward brainly !!

Reil [10]

What happens to the water in the clouds is that he cloud gets heavy and let’s the rainfall out, aka rain

8 0

3 years ago

If you want to make 8.00 moles of AlF₃ how many moles of F₂ will you need, using the following balanced chemical equation? 2 Al

Zinaida [17]

Answer:

You will need 12 moles of F2 if you want to make 8 moles of AlF3.

Explanation:

It takes 3 moles F2 to make 2 moles of AlF3 (this will be our mole ratio)

2 moles AlF3/3 moles F2 =8 moles AlF3/x moles AlF3.

x=12 moles AlF3

7 0

2 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago