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VladimirAG [237]
3 years ago
9

Please help actual answer

Chemistry
1 answer:
Fed [463]3 years ago
4 0

Answer:

Explanation:

E.92

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Calculate the amount (in grams) of kcl present in 75.0 ml of 2.10 m kcl
Lera25 [3.4K]
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol

74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl

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5 0
3 years ago
What are some differences between an “ideal” gas and real gases?
gogolik [260]

Answer:

The Ideal Gases are described under the standard conditions to eliminate as many variables as possible.

Explanation:

P*V = n*R*(temp final-temp initial) There are standard conditions for the relationships between pressure (P), Volume (V), number of moles (n), and the temperature in Kelvin (K) or Celsius (C).

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How are the processes of erosion and deposition closely related
denpristay [2]

First of all, hey there! I hope you’re having a good day.

They are related because erosion moves broken materials and deposition means that the material is getting deposited, so that’s kind of how they are related. I hope you understand!!:)

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3 0
3 years ago
Substance A is a gas, Substance B is a liquid and Substance C is a solid. All of the substances are at the same temperature and
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Explanation:

4 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)

Substance      ΔG°f (kJ/mol)

M₃O₄                 -9.50

M(s)                       0

O₂(g)                     0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

PCl_3(l)\rightarrow PCl_3(g)

First we have to calculate the Gibbs energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]

where,

\Delta G^o = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]

\Delta G^o=9.50kJ/mol

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs free energy  = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K  = equilibrium constant = ?

9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)

K=0.0216

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

K=(P_{O_2})^2

0.0216=(P_{O_2})^2

P_{O_2}=0.147atm

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0
3 years ago
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