What ion is formed when an atom of magnesium (mg) loses two electrons? mg–1 mg

According to vsepr theory the molecular geometry for ch3+ is

CaHeK987 [17]

According the VSEPR theory the molecular geometry for CH3+ is triagonal planar

8 0

3 years ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

Be sure to answer all parts.

Llana [10]

Answer:

16 g has 2 sig figs

6701g has 4 sig figs

560 has 2 sig figs

Explanation:

4 0

2 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

2 years ago

What ion is formed when an atom of magnesium (mg) loses two electrons? mg–1 mg–2 mg 1 mg 2