The ionic charge of an element in Group 6A is 2

1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u

Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.

1) We can find the next equation:
mass=molar mass x number of moles.

x=number of moles of helium
y=number of moles of helium.

(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:

(1)

4x+83.8y=103.75

2) We can find other equation:

We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.

1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).

Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.

x=30% of (x+y)

Therefore, we have this other equation:
(2)

x=0.3(x+y)

With the equations(1) and (2), we have the next system of equations:

4x+83.8y=103.75

x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7

We solve this system of equations by substitution method.
x=3y/7

4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21

x=3y/7=3(1.21)/7=0.52

We have 0.52 moles of helium and 1.21 moles of Krypton.

1 mol=6.022 * 10²³ atoms

Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).

Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.

Answer: The sample have 1.044 * 10²⁴ atoms.

6 0

4 years ago

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Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.

Sindrei [870]

The molar solubility is 7.4× $10^{-3}$ M and the solubility is 7.4× $10^{-3}$ g/L .

Calculation ,

The dissociation of silver bromide is given as ,

$AgBr$ → $Ag ^{+}$ + $Br^{-}$

S

- S S

Ksp = [ $Ag ^{+}$ ] [ $Br^{-}$ ] = [S] [ S ] = $S^{2}$

S = √ Ksp = √ 5. 5× $10^{-5}$ = 7.4× $10^{-3}$

The solubility =7.4× $10^{-3}$ g/L

The molar solubility is the solubility of one mole of the substance.

Since , one mole of $AgBr$ is dissociates and form one mole of each $Ag ^{+}$ and $Br^{-}$ ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4× $10^{-3}$ mol/L = 7.4× $10^{-3}$ M

To learn more about molar solubility ,

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7 0

1 year ago

1. How many moles are in 3.4 x 1024 molecules of HCl?

pochemuha

1. How many moles are in 3.4 x 10²⁴ molecules of HCl?

if there are 6.022 × 10²³ molecules in 1 mole of HCl

then there are 3.4 × 10²⁴ molecules in X moles of HCl

X = (3.4 x 10²⁴ × 1) / 6.022 × 10²³ = 5.6 moles of HCl

2. How many grams are in 2.59 x 10²³ formula units of Al₂O₃?

if there are 6.022 × 10²³ units in 1 mole of Al₂O₃

then there are 2.59 × 10²³ units in X moles of Al₂O₃

X = (2.59 x 10²³ × 1) / 6.022 × 10²³ = 0.43 moles of Al₂O₃

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of Al₂O₃ = 0.43 × 102 = 43.86 g

3. How many grams are in 9.05 x 1023 atoms of silicon?

if there are 6.022 × 10²³ atoms in 1 mole of silicon

then there are 9.05 × 10²³ atoms in X moles of silicon

X = (9.05 x 10²³ × 1) / 6.022 × 10²³ = 1.5 moles of silicon

mass = number of moles × molecular weight

mass of silicon = 1.5 × 28 = 42 g

4. If you had 7.00 moles of PtO₂, how many grams would you have?

mass = number of moles × molecular weight

mass of PtO₂ = 7 × 227 = 1589 g

5. How many moles are in 29.2 L of oxygen gas at STP?

At standard temperature and pressure (STP) we may use the following formula:

number of moles = volume (L) / 22.4 (L / mol)

number of moles of oxygen = 29.2 / 22.4 = 1.3

6. What is the volume (liters) of 75.8 g of N₂ at STP?

number of moles = mass / molecular weight

number of moles of N₂ = 75.8 / 28 = 2.7 moles

At standard temperature and pressure (STP) we may use the following formula:

number of moles = volume (L) / 22.4 (L / mol)

volume = number of moles × 22.4

volume of N₂ = 2.7 × 22.4 = 60.48 L

4 0

4 years ago

The ionic charge of an element in Group 6A is 2–.