How you will devide the meterial based on their solubility

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago

In the ground state, an atom of which element has seven valence

sweet-ann [11.9K]

Answer:

Fluorine

Explanation:

Fluorine has 9 total electrons. The first two are in the 1s level, and the remaining electrons are on the outer level of the atom, with 2 in the s level and 5 in the p level. The electron configuration is 1s2 2s2 2p5.

7 0

3 years ago

What is the mass of 2.25 x 10^25 atoms of lead

NemiM [27]

Answer: the amount of a substance that contains 6.02 x 1023 respective particles of that

substance

Avogadro’s number: 6.02 x 1023

Molar Mass: the mass of one mole of an element

CONVERSION FACTORS:

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Try:

1. How many atoms are in 6.5 moles of zinc?

6.5 moles 6.02 x 1023 atoms = 3.9 x 1024 atoms

1 mole

2. How many moles of argon are in a sample containing 2.4 x 1024 atoms of argon?

2.4 x 1024 atoms of argon 1 mole = 4.0 mol

6.02 x 1023 atoms

3. How many moles are in 2.5g of lithium?

2.5 grams Li 1 mole = 0.36 mol

6.9 g

4. Find the mass of 4.8moles of iron.

4.8 moles 55.8 g = 267.84 g = 270g

1 mole

MOLE PARTICLES

(ATOM)

MASS

(g)

1 mole = molar

mass

(look it up on

the PT!)

1 mole =

6.02 x 1023

atoms

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Two Step Problems:

1. What is the mass of 2.25 x 1025 atoms of lead?

2.25 x 1025 atoms of lead 1 mole 207.2g = 7744.19g = 7740g

6.02 x 1023 atoms 1 mole

2. How many atoms are in 10.0g of gold?

10 g gold 1 mole 6.02 x 1023 atoms = 3.06 x 1022 atoms

197.0g 1 mole

PRACTICE PROBLEMS:

1. How many moles are equal to 625g of copper?

625g of copper 1 mol = 9.77 mol Cu

64 g Cu

2. How many moles of barium are in a sample containing 4.25 x 1026 atoms of barium?

4.25 x 1026 atoms of barium 1 mol = 706 mol

6.02 x 1023 atoms

3. Convert 2.35 moles of carbon to atoms.

2.35 moles 6.02 x 1023 atoms = 1.41 x 1024 atoms

1 mole

4. How many atoms are in 4.0g of potassium?

4.0 g 1 mole 6.02 x 1023 atoms = 6.2 x 1022 atoms

39.1 g 1 mole

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

5. Convert 9500g of iron to number of atoms in the sample.

9500 g Fe 1 mole 6.02 x 1023 atoms = 1.0 x 1026 atoms

55.8g 1 mole

6. What is the mass of 0.250 moles of aluminum?

0.250 moles 27.0g = 6.75 g Al

1 mole

7. How many grams is equal to 3.48 x 1022 atoms of tin?

3.48 x 1022 atoms 1 mole 118.7g = 6.86 g Sn

6.02 x 1023 atoms 1 mole

8. What is the mass of 4.48 x 1021 atoms of magnesium?

4.48 x 1021 atoms 1 mole 24.3g = 0.181 g Mg or 1.8 x 10-1

6.02 x 1023 atoms 1 mole

9. How many moles is 2.50kg of lead?

2.50kg 1000 g 1 mol = 12.1 mol

1 kg 207.2 g

10.Find the mass, in cg, of 3.25 x 1021 atoms of lithium.

3.25 x 1021 atoms 1 mol 6.9g 100cg = 3.7

Explanation:

5 0

3 years ago

Identify the species oxidized, the species reduced, the oxidizing agent and the reducing agent in the following electron transfe

babunello [35]

Answer:

Oxidized and reducing agent: manganese.

Reduced and oxidizing agent: mercury.

Explanation:

Hello!

In this case, for the reaction:

$Hg^{2+}(aq) + Mn^0(s)\rightarrow Hg(s) + Mn^{2+}(aq)$

We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.

Best regards

6 0

2 years ago

Suppose of hcl and of o2 are added to an empty flask. How much will be in the flask at equilibrium?A. None. B. Some, but less th

serg [7]

This is an incomplete question, here is a complete question.

Hydrogen chloride and oxygen react to form water and chlorine, like this:

$4HCI(g)+O_2(g)\rightarrow 2H_2(g)+2CI_2(g)$

Use this chemical equation to answer the questions in the table below.

Suppose 155. mmol of HCl and 38.8 mmol of O₂ are added to an empty flask. How much HCl will be in the flask at equilibrium?

(1) None.

(2) Some, but less than 155. mmol,

(3) 155. mmol,

(4) More than 155. mmol,

Answer : The correct option is, (1) None

Explanation :

The given balanced chemical reaction is:

$4HCI(g)+O_2(g)\rightarrow 2H_2(g)+2CI_2(g)$

From the balanced chemical reaction, we conclude that

As, 4 mmole of HCl react with 1 mmoles of $O_2$

So, 155. mmole of HCl react with $\frac{155.}{4}\times 1=38.75$ mmoles of $O_2$

That means, $O_2$ is excess reagent and $HCl$ is a limiting reagent. All the moles of HCl will consume.

Hence, the correct option is, None.

7 0

3 years ago

How you will devide the meterial based on their solubility​

How you will devide the meterial based on their solubility