All categories

3 years ago

1. What would be the molarity of the sodium ion in solution.

1 answer:

5 0

Therefore, the molar concentration of sodium ions is just 1 TIMES the molar concentration of sodium chloride. 1 times 0.110 equals 0.110. Therefore: [Na+] = 0.110. Na+ with square brackets around it is read as "molar concentration of Na+".

You might be interested in

How many cups of beer are in a keg?

Alinara [238K]

The answer is roughly 165 cups of beer.

5 0

3 years ago

Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative

masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

3 0

3 years ago

To separate a mixture of iron filings and salt, the most efficient method would be

GaryK [48]

Answer:

Explanation:

Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.

I do hope I helped you! :)

4 0

3 years ago

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

nignag [31]

Answer: 1) Maximum mass of ammonia 198.57g

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g

Explanation:

In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0

3 years ago

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

6 0

3 years ago

1. What would be the molarity of the sodium ion in solution.​

1. What would be the molarity of the sodium ion in solution.