Which choice(s) include(s) d-orbital contribution in the hybridization scheme: pcl3, no3─, i3─, h2se ?

Answers

Answer 1

Answer:

Among all three PCl₃, NO₃⁻ , I₃⁻, H₂Se only I₃⁻ will involve participation of d-orbitals in hybridization as bonding in I₃⁻ will include, s, p and d orbital as shown in the image attached. there will be sp³d hybridization, there will be presence of three lone pairs and 2 bonds as

I⁻ has 8 valence electrons and 2 neighbours atoms which will need one electron each to satisfy their valency.

so the number of electrons on central atom will be:

8-1-1=6

That 6 electrons will make 6/2 =3 lone pairs.

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One way to experimentally measure the heat capacity of a Styrofoam cup calorimeter would be to melt a known mass of ice in warm water and measure hte temperature change. Use the data below to determine the experimental heat capacity of the calorimeter. Use the literature heat of fusion for ice in your calculations. Assume the ice added is at 0.00 c.Mass of ice added: 17.69gMass of water in calorimeter: 98.67gT-Initial of water: 28.7T-Final of water after melting ice: 12.9C

Answers

Answer:

4.88 Cals per degree celsius

Explanation:

We have taken heat of fusion of ice = 80 cals / g

We have taken speciic heat of water = 1 cal/g per degree celsius

In this experiment , let the heat capacity of calorimeter be X.

Heat gained by ice

heat gained in melting + heat gained in getting warmed

= mass x latent heat + mass x specific heat x rise in temperature

= 17.69 x 80 + 17.69 x 1 x ( 12.9 - 0 )

= 1643.4 Cals

Heat lost by water

= mass x specific heat x fall in temperature

98.67 x 1 x ( 28.77 - 12.9 )

= 1565.89 Cals

Heat lost by calorimeter

heat capacity x fall in temperature

X x ( 28.77 - 12.9 )

Heat gained = heat lost

1643.4 = 1565.89 +15.87X

X = 4.88 Cals per degree celsius

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

Answers

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

Learn more about titration here:

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Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mLmL of water to produce a solution that freezes at −−14.5 ∘C? The freezing point for pure water is 0.0 ∘C∘C and Kf is equal to 1.86 ∘C/m.If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

Answers

Answer:

1) 108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) 1.73 is the actual value of the van't Hoff factor, i.

Explanation:

1) Formula used depression in freezing point ;

$\Delta T_f=T-T_f$

$\Delta T_f=i* K_f* m$

where,

$T_f$ =Freezing point of solution

T = Freezing point of water

$\Delta T_f$ =depression in freezing point =

i = van't Hoff factor of solute

$K_f$ = freezing point constant

m = molality of solution

We have :

$K_f$ of water = 1.86°C/m ,

Molality of solution = m = ?

$KNO_3(aq)\rightarrow K^+(aq)+NO_3^(-)(aq)$

i = 2

Freezing point of solution = $T_f=-14.5^oC$

Freezing point of pure water = T = 0°C

$\Delta T_f=T-T_f$

$\Delta T_f=0^oC-(-14.5 ^oC)=14.5^oC$

$14.5^oC=2* 1.86^oC* m$

m = 3.898 molal

3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water.

Volume of water , V= 275 ml

Mass of water = m

Density of water= d = 1 g/mL

$m=d* V=1 g/ml* 275 mL = 275 g$

Here, 3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water. Then moles of potassium nitarte present in 275 grams of water is :

$(3.989 mol)/(1000)* 275 =1.072 mol$

Mass of 1.072 moles of potassium nitrate :

1.072 mol × 101 g/mol = 108.27 g

108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) Formula used an Elevation in boiling point;

$\Delta T_b=T_b-T$

$\Delta T_b=i* K_b* m$

where,

$T_b$ =boiling point of solution

T = boiling point of water

$\Delta T_b$ =Elevation in boiling point =

i = van't Hoff factor of solute

$K_b$ = Boiling point constant

m = molality of solution

of the solution

We have :

$K_b$ of water = 0.512°C/m ,

Molality of solution = m = 3.90 m

i =?

The boiling point of pure water = T = 100.00°C

The boiling point of solution = $T_b$ = 103.45°C

$\Delta T_b=103.45^oC-100.00^oC=3.45^oC$

$\Delta T_b=i* K_b* m$

$3.45^oC=i* 0.512 ^oC/m* 3.90 m$

$i=(3.45^oC)/(0.512 ^oC/m* 3.90 m)=1.73$

1.73 is the actual value of the van't Hoff factor, i.

An 18-karat gold necklace is 75% gold by mass, 16% silver, and 9.0% copper.A. What is the mass, in grams, of the necklace if it contains 0.24 Oz of silver?
B. How many grams of copper are in the necklace.?
C. If the 18- karat gold has a density of 15.5g/cm^3, what is the volume in cubic centimeters?

Answers

B. How many grams of copper are in the necklace.?

I think that the answer would be option B.

Can I+ (the iodine cation) becalled a Lewis base?
Fullyexlplain your answer.

Answers

Answer: No I+ cannot be called a Lewis base.

Explanation:

According to Lewis Theory, it defines an acid as an electron-pair acceptor and a base as an electron-pair donor.

In terms of Lewis basicity, Iodide ion (anions) has the more readily available lone pair electrons for donation since iodide ion is less electronegative .

With the help of the net electronic structure one can understand the answer of the question, because we need to study the I+ ion (cation) structure.

Lewis acid is therefore any substance, that can accept a pair of nonbonding electrons.

From the picture below I+ is most likely ready to accept electrons not to give from it 5s orbital to become stable.

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.