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Lisa [10]
3 years ago
6

Please someone do this for me. Im really stuck

Chemistry
2 answers:
Alex_Xolod [135]3 years ago
7 0
Is it me or is the picture not loading
Alina [70]3 years ago
5 0

HOLY COW , Im taking the same exact class , online adv. Chem? Which lab are you doing i will send it to you. Add me on the snap too andyboi2001

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How does the attraction between particles affect the ability of a solvent to dissolve in a substance
Anni [7]

Explanation:

The attraction between solute particles affect its ability to dissolve in the solution because more is the attraction between particles less will be the chances that solute particles will combine to the solvent particles.

As a result, there will be decrease in solubility with increase in force of attraction between solute particles.

Whereas when there is less force of attraction between solute particles then more easily solute particles will combine with the solvent particles.

Hence, with decrease in force of attraction between solute particles, the rate of dissolution will increase.

6 0
3 years ago
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The angle formed by the nuclei of two surrounding atoms with the nucleus of the central atom in a structure is called a(n) _____
yulyashka [42]

The bond angle obtained based on the geometry of the molecule alone is known as the predicted bond angle.

When a compound is formed, the less electronegative atom in the compound is called the central atom in the molecule. The angle formed by the nuclei of two surrounding atoms with the nucleus of the central atom in a structure is called a bond angle.

The bond angle is usually based on the geometry of the molecule. The expected geometry can be deviated for certain reasons such as the presence of lone pairs on the central atom in the molecule. The bond angle obtained based on the geometry of the molecule alone is known as the predicted bond angle.

Learn more about bond angle: brainly.com/question/6179102

5 0
3 years ago
how many electrons are transferred between the cation and anion to form the ionic bond in one formula unit of each compound? (1
Elina [12.6K]

Answer:

The answer to your question is:

Explanation:

Compound                 Cation                 Anion         Number of electrons

LiCl                              Li⁺¹                        Cl⁻¹                  one

NaF                             Na⁺¹                       F⁻¹                    one

CaO                            Ca⁺²                       O⁻²                  two

BaS                             Ba⁺²                        S⁻²                  two

NaBr                           Na⁺¹                       Br⁻¹                   one

8 0
3 years ago
Read 2 more answers
At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
Calculate the volume of .987 mol of gas at stp will occupy
aalyn [17]
Number of moles = volume / (molar volume)
Molar volume at stp = 22.4 dm^3
Volume = no of moles × molar volume
= 0.987 × 22.4
= 22.1088 dm^3
= 22108.8 cm^3

Hope it helped!
5 0
3 years ago
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