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3 years ago

What does point of view refer to?

Chemistry

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

I'd say the most reasonable answer is perspective.

Explanation:

"Point of view" is about who is telling and sharing the details in what you are writing or reading. I'm not 100% sure because the word "perspective" doesn't seem like the right word to use, but it's better than the other two in my opinion.

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When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio

arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)

ΔT = 13.4°C - (-12.82) = 26.22°C

m is molality of the solution

Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)

And i is Van't Hoff factor (1 for all solutes in benzonitrile)

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.

5 0

2 years ago

A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

3 0

3 years ago

Which compound is a carbohydrate. Nucleotide or protien or starch or amino acid

marshall27 [118]

Starch is carbohydrate

3 0

3 years ago

If equal amounts of silver ion are added to silver bromide and silver chloride, why does the bromide mixture precipitate first?

VARVARA [1.3K]

Answer is: precipitation requires fewer Ag⁺ ions in AgBr than in AgCl.
Chemical reactions:
Ksp(KBr) = 5,3·10⁻¹³.
Ksp(KCl) = 1,8·10⁻¹⁰.
Ksp is solubility product constant. The higher the Ksp value, substance is more soluble. KBr has lower Ksp, so it is easier to form precipitant of KBr than KCl.

5 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$