Strontium oxide + ? = stronium chloride

Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

4 0

3 years ago

what happens to the average kinetic energy of ideal gas molecules when the pressure of the gas is increased by reducing the volu

Margarita [4]

Answer:

The pressure of the gas is increased by reducing the volume at constant temperature. ... The average velocity of the molecules is increased by a factor of 2.

Explanation:

4 0

2 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

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8 0

1 year ago

Label the liquids from least to most dense, Vegetable oil, water, Dish soap, Corn syrup, Honey

masya89 [10]

Answer:

1. Vegetable oil

2. Water

3. Dish soap

4. Corn syrup

5. Honey

8 0

3 years ago

What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of 285K?

julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

$\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm$

7 0

3 years ago

Strontium oxide + ? =stronium chloride​

Strontium oxide + ? =stronium chloride