Balance the equation: KL+Pb(NO3)2→PbL2+KNO3

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Answer 1

Answer: is that Lr? I'm not sure what the L one stand for?
2:1:1:2

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When the chemical equation is balanced what is the ratio of the coefficient of Ag(s) to the cofficient of H2O(l)?___Ag(s)+____NO3-(aq)+____H+(aq)â†’ ___Ag+(aq)+____NO(g)+______H2O(l)a. 1:1b. 2:1c. 3:1d. 3:2
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a student wants to remove the salt from a mixture of sand and salt in order to get only person he adds water to the mixture why it's the step help

Answers

You would add water to the mixture to dissolve the salt. When the salt is dissolved, you have your sand alone.

Water dissolves the salt and pour out the water let it evaporate and you have separate salt and water

For the experiments from 1-3 with the same temperature change, what other parameters are the same at the initial reaction conditions?a. mmol HCl total mL solution
b. mmol NaOH Initial concentration of NaOH
c. mL H2O Initial concentration of HCl

Answers

Answer:

The following parameters:

total mL solution
mmol NaOH
initial concentration of NaOH

Explanation:

The options have been incorrectly arranged in the question statement given here. The three conditions mentioned above are what would be observed to remain the same when this experiment is carried out.

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value

Answers

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

$Ksp=[Ag^+][Cl^-]$

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

I - 0 0

C - +x +x

E - x x

Which leads to the following modified equilibrium expression:

$Ksp=x^2$

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached New England, the rain it produced had an average pH of 4.66. How much more acidic was the rain falling in New England?

Answers

Answer:

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

Explanation:

The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.

pH = -log ([H⁺])

American Midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

New England

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

The ratio of concentrations is:

$(2.19 * 10^(-5) M )/(9.55 * 10^(-6) M) =2.29$

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

The change in the acidity of the rain falling in New England is mathematically given as

r=2.29

What is the change in the acidity of the rain falling in New England?

Question Parameter(s):

American midwest pH of 5.02

New England pH of 4.66

Generally, the equation for the pH is mathematically given as

pH = -log ([H⁺])

Therefore American midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

and New england

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

In conclusion, the concentration ratio is

$r=(2.19 * 10^(-5) M )/(9.55 * 10^(-6) M)$

r=2.29

Answers

The element that is found in all organic compounds is Carbon

A weather balloon is inflated to a volume of 27.3 L at a pressure of 738 mmHg and a temperature of 26.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -15.6 ∘C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answers

Answer:

The new volume of the balloon is 46.1 L

Explanation:

Step 1: Data given

Initial volume of the balloon = 27.3 L

Initial pressure in the balloon = 738 mmHg = 0.97105 atm

Initial temperature in the balloon = 26.9 °C = 300.05 K

The pressure decreases to 375 mmHg = 0.493421 atm

The temperature lowers to -15.6 °C = 257.55 K

Step 2: Calculate the volume

P1*V1 / T1 = P2 *V2 / T2

⇒with P1 = the Initial pressure = 738 mmHg = 0.97105 atm

⇒with V1 = Initial volume of the balloon = 27.3 L

⇒with T1 = Initial temperature in the balloon = 26.9 °C = 300.05 K

⇒with P2 = the decreased pressure = 0.493421 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the lowered temperature = 257.55 K

0.97105 * 27.3 / 300.05 = 0.493421*V2 / 257.55

V2 = 46.1 L

The new volume of the balloon is 46.1 L

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