Any two objects that have Mass will also have which type of force between them

Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no rea

iragen [17]

Answer :

(A) The balanced molecular equation will be:

$K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)$

(B) The balanced molecular equation will be:

$Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)$

(C) The balanced molecular equation will be:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

(D) The balanced molecular equation will be:

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}$

Explanation :

Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

Part A : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

$K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)$

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

Part B : lithium sulfate and lead(II) acetate

The balanced molecular equation will be:

$Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)$

In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.

Part C : copper(II) nitrate and sodium sulfide

The balanced molecular equation will be:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.

Part D : strontium nitrate and potassium iodide

The balanced molecular equation will be:

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+SrI_2(aq)$

In this reaction, strontium iodide and potassium nitrate are soluble solution.

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}$

6 0

3 years ago

When 1 mol of KBr(s) decomposes to its elements, 394 kJ of heat is absorbed. (a) Write a balanced thermochemical equation.

VladimirAG [237]

The balanced thermochemical equation is

KBr ------- K + 1/2 Br2

<h3>What is thermochemical equation? </h3>

A Thermochemical Equation is defined as the balanced stoichiometric chemical equation which includes the enthalpy change, ΔH.

The chemical equation for the decomposition of potassium bromide to its constituent elements bromine ans potassium :

KBr ----- K + Br2

The balanced thermochemical equation of the decomposition of potassium bromide to its constituent elements potassium and bromide as follows

KBr ------- K + 1/2 Br2

As the heat is absorbed in this reaction therefore, heat is positive.

Thus, we concluded that the balanced thermochemical equation is

KBr ------- K + 1/2 Br2

learn more about thermochemical equation:

brainly.com/question/2733624

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8 0

1 year ago

What force makes a ball rolling along the ground slow down

Mashcka [7]

Friction is the force that slows down a ball, if their were no friction, it would not stop.

Answer: friction

6 0

3 years ago

Using the idea that reactions occur as a result of collisions between particles, explain why reaction rates depend on the temper

PIT_PIT [208]

Increasing temperatures in a reaction increases the kinetic energy of the reactant molecules. This causes them to move fast and hence collide with a higher frequency. The higher the rate of collision between the molecules, the faster the reaction.

6 0

2 years ago

A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure

GenaCL600 [577]

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

Best regards!

8 0

2 years ago