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3 years ago

Suppose you have a mixture of solid mg(oh)2 and baso4, how do you separate the mixture

1 answer:

5 0

This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.

Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O

The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.

MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄

Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.

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A chlorine atom in its ground state has a total of seven electrons in orbitals related to the atoms third energy level. how many

kiruha [24]

5 electrons

The electron configuration of an atom shows or describes the distribution of electrons in their atomic orbitals. In particular, it shows the distribution in terms of the principal energy levels or electron shells (reflected by the numbers 1, 2, 3, 4, 5, 6, 7) and their corresponding subshells (reflected by the letters s,p,d,f,g) composed of atomic orbitals.

The maximum number of electrons in the s-subshell is 2, in the p-subshell, 6, in the d-subshell, 10, in the f-subshell, 14, and in the g-subshell, 18. In the electron configuration of an atom, the number of electrons in the subshell for a given atom is reflected by a number (in superscript) next to the corresponding letter of the subshell.

The electron configuration of a chlorine atom in its ground state is
 $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$

Thus, the number of electrons in the 3rd energy level in the p-subshell is 5 ( $3p^{5}$ as seen in the electron configuration of the atom).

7 0

3 years ago

PLEASE HELP ME!!

Mama L [17]

Answer:

Explanation:

3 CARBON ATOMS 8 HYDROGEN

5 0

3 years ago

13. An aerosol spray can of deodorant with a volume of 0.410 L contains 3.0 g of propane gas (CH3) as propellant. What is the pr

qaws [65]

Answer: The pressure in the can is 4.0 atm

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = ?

V = Volume of gas = 0.410 L

n = number of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{3.0g}{44.1g/mol}=0.068mol$

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $20^0C=(20+273)K=293K$

$P=\frac{nRT}{V}$

$P=\frac0.068mol\times 0.0820 L atm/K mol\times 293K}{0.410L}=4.0atm$

Thus the pressure in the can is 4.0 atm

8 0

3 years ago

How many milliners of hydrogen gas ar produced by the reaction 0.020 moles of magnesium with excess of hydrochloride acid at sto

jeyben [28]

Answer:- 448 mL of hydrogen gas are formed.

Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

$Mg+2HCl\rightarrow MgCl_2+H_2(g)$

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.

moles of Hydrogen gas formed = 0.020 mol

At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

$0.02mol(\frac{22.4L}{1mol})$

= 0.448 L

They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

$0.448L(\frac{1000mL}{1L})$

= 448 mL

So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.

5 0

3 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

3 years ago