What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!

Calculate the molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water. Use 7.6 x 10^-36 as the solubility product constant

Molodets [167]

Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

3Cu2+ + 2(AsO4)3- = Cu3(AsO4)2

7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M

8 0

3 years ago

Read 2 more answers

Why do we put these three words together in the 'fire triangle'?

bazaltina [42]

The three word of this is to represent what it takes to create and maintain a fire

7 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

1. Name three obstacles that Mendeleev faced in his life?

Fittoniya [83]

Answer:

was difficult to place isotopes of elements as they have the same chemical properties but different atomic masses. It was not possible to predict how many elements could be discovered between two heavy elements as the rise in atomic mass is not uniform.

3 0

3 years ago

The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitr

timama [110]

1) number of moles of N2 = n/2

2) Number of moles of CH4 = n/2

3) Total number of moles of the mixture = n/2 + n/2 = n

4) Kg of N2

mass in grams = number of moles * molar mass

molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol

=> mass of N2 in grams = (n/2) * 28 = 14n

mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg

Answer: mass of N2 in kg = 0.014n kg

3 0

3 years ago