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2 years ago

Convert 2.5 X 10^24 atoms of copper to grams of copper

Chemistry

1 answer:

viva [34]2 years ago

7 0

2.5 X 10²⁴ atoms of copper is equivalent to 266.7g of Cu.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;

mass of substance (g) = no. of moles (mol) × molar mass (g/mol)

However, the number of moles of copper must first be calculated by dividing the number of atoms by Avogadro's number as follows:

no. of moles of Cu = 2.5 X 10²⁴ ÷ 6.02 × 10²³

no. of moles = 2.5/6.02 × 10²⁴-²³

no. of moles = 0.42 × 10¹

no. of moles = 4.2moles of copper.

Molar mass of copper = 63.5g/mol

Mass of copper = 4.2mol × 63.5g/mol

Mass of copper = 266.7g

Therefore, 2.5 X 10²⁴ atoms of copper is equivalent to 266.7g of Cu.

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3 years ago

What is the isotope notation for an ion of silver-109 with a charge of positive 1

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Answer:

The isotope notation for an ion of silver-109 with a charge of positive 1 is $_{47}^{109}Ag^{1+}$ .

Explanation:

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From the above two isotopes have same atomic number and different mass number.

From the given,the isotope notation for an ion of silver-109 with a charge of positive 1

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2 years ago

Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w

jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, $COCl_2$ , gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

The value of $K_c$ for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which $p_{CO}=p_{Cl_2}=0.265atm$ and $p_{COCl_2}=0.000atm$ ?

<u>Answer:</u> The equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of $K_c\text{ and }K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure

$K_c$ = Equilibrium constant in terms of concentration = 5.79

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side = $n_{g,p}-n_{g,r}=1-2=-1$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = Temperature = 570 K

Putting values in above equation, we get:

$K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124$

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

<u>Initial:</u> 0.265 0.265

<u>At eqllm:</u> 0.265-x 0.265-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}$

Putting values in above equation, we get:

$0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59$

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = $(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $Cl_2=(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $COCl_2=x=0.008atm$

Hence, the equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

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3 years ago

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<h3>What is molarity?</h3>

The molarity of a solution is defined as the concentration of a solute per unit volume of the solution.

Given, the volume of the solution is 250 ml converted into l 0.25

The number of moles of NaCl is 0.70.

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2 years ago