Please help with these two (last question is nwse)

ICl has a higher boiling point than Br2. What is the best explanation for this?

Flura [38]

The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.

7 0

3 years ago

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

Answer: The molality of the solution is 0.11 m

Explanation:

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

Convert 16.8 L of nitrogen monoxide gas at STP to grams .

german

Answer:

See explanation

Explanation:

1 mole of a gas occupies 22.4 L

x moles occupies 16.8 L

x = 1 mole * 16.8 L/22.4 L

x = 0.75 moles

number of moles = mass/molar mass

mass = number of moles * molar mass

mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g

the coefficient of the scientific notation answer = 2.25

the exponent of the scientific notation answer = 1

significant figures are there in the answer = 6

the right most significant figure in the answer = 3

2.

number of moles = 12.5g/38g/mol = 0.3289 moles

1 mole occupies 22.4 L

0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole

= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L

the coefficient of the scientific notation answer =7.37

the exponent of the scientific notation answer = 0

significant figures are there in the answer = 6

the right most significant figure in the answer= 3

8 0

3 years ago

When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate f

Dahasolnce [82]

KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which then will react with 4-nitroaniline to form a diazonium salt.

8 0

3 years ago

a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what

e-lub [12.9K]

The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K

p_1 = 1014 kPa

p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa

5 0

3 years ago