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vitfil [10]
2 years ago
5

What is the name and formula of the compound formed between lithium and sulfite?

Chemistry
1 answer:
dmitriy555 [2]2 years ago
6 0

Answer:

d)Lithium sulfite, Li2SO3

Explanation:

LITHIUM is an element in the group 1 (alkali metals) of the periodic table, with a charge of +1. On the other hand, SULFITE is a radical with the chemical formula (SO3²-). It is made up of sulfur and oxygen atoms.

According to this question, lithium element and sulfite radical combines to form a compound called LITHIUM SULFITE. From the chemical formula of these individual constituents, the chemical formula of the compound formed will go this: Li2SO3. This means that it takes 2 atoms of Li+ to combine with SO3²- in order to form a neutral compound.

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If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a 1.00-L flask and t
Furkat [3]

Answer:

The resulting pressure in the flask is 0.93 atm

Explanation:

- Apply the Ideal Gas law in both cases to get the mols, of Ar and SO2.

- Once you know the mols, sum both to get the total mols in the mixture.

- Apply the Ideal Gas lawin the flask with the total mols to know the resulting pressure.

First: 0.750 L of argon at 1.50 atm and 177°C

T° C + 273 = T° K → 177°C + 273 = 450K

P .V = n . R . T

1.50 atm . 0.750 L = n . 0.082 L.atm/mol.K  . 450K

(1.50 atm . 0.750 L) /  (0.082 mol.K/L.atm  . 450K) = n

0.030 mols Ar = n

Be careful with the R units, the ideal gases constant

Let's convert kPa to atm.

101.33 kPa _____ 1 atm

95 kPa ________ (95 / 101.33) = 0.94 atm

T° C + 273 = T° K → 63°C + 273 = 336 K

0.94 atm . 0.235 L = n . 0.082 L.atm/mol.K . 336K

(0.94 atm . 0.235 L) / (0.082 mol.K/L.atm . 336K) = n

8.01X10⁻³ mols = n

0.030 mols Ar  + 8.01X10⁻³ mols SO₂ = 0.038 total mols in the mixture

1L . P = 0.038 mol . 0.082 L.atm / mol.K . 298 K

P = (0.038 mol . 0.082 L.atm / mol.K . 298 K ) / 1L

P = 0.93 atm

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8 0
3 years ago
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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
Read 2 more answers
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