The uppermost layer of the atmosphere

- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (Ag

RideAnS [48]

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is $AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)$

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

3 0

2 years ago

What determines an element's chemical properties

Alik [6]

The three factors determine the chemical properties of an element:
The number and arrangement of electrons in an atom
The number of valence electrons
The number and arrangement of electrons

5 0

2 years ago

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Consider a balloon with volume V. It contains n moles of gas and has an internal pressure of P. The temperature of the gas is T.

Maksim231197 [3]

Answer:

V₂ = 0.6 V.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n is constant, and have different values of P, V and T:

(P₁V₁T₂) = (P₂V₂T₁).

V₁ = V, P₁ = P, T₁ = T.

V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.

∴ V₂ = (P₁V₁T₂)/(P₂T₁) = (P)(V)(0.75 T)/(1.25 P)(T) = 0.6 V.

3 0

2 years ago

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Consider the reaction.

Alex_Xolod [135]

Note the signs of equilibrium:-

Reaction don't procede forward or backward
Concentration of products and reactants remains same .

So

if

Concentration of A is 2M then concentration of B should be same .

So equilibrium constant K is 1

$\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}$

So

[B]=[A]^2
[B]=2^2
[B]=4M

8 0

1 year ago

aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is

maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0

2 years ago