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2 years ago

Please write balanced equations for these reactions! WILL GIVE BRAINLIEST! 50 POINTS

Chemistry

1 answer:

Romashka [77]2 years ago

3 0

Answer:

(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)

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Which metalloid has three valence electrons? 1. Lithium 2. boron 3. silicon 4. aluminum 5. arsenic

iren [92.7K]

Answer:

boron

Explanation:

5 0

3 years ago

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Which dilute acid is used to make sodium carbonate salt

andreev551 [17]

Hydrochloric acid :))

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2 years ago

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A 4.00g sample of helium has a volume of 24.4L at a temperature of 25.0oC and a pressure of 1.00 atm. The volume of the helium i

Masteriza [31]

Answer:

0.41 moles.

Explanation:

Given that:

Mass of helium = 4.00 g

Initial Volume = 24.4 L

initial Temperature = 25.0 °C =( 25 + 273) = 298 K

initial Pressure = 1.00 atm

The volume was reduced to :

i.e

final volume of the helium - 10.4 L

Change in ΔV = 24.4 - 10.4 = 10.0 L

Temperature and pressure remains constant.

The new quantity of gas can be calculated by using the ideal gas equation.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{1.00*10.0}{0.082057*298}$

n = 0.4089 moles

n = 0.41 moles.

7 0

2 years ago

Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume

DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

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For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

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<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

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Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

4 0

2 years ago

If two atoms are bonded with a bond energy of 87 kJ, what must be done to break the bond?

SSSSS [86.1K]

I would say the energy has to be decreased by 87 kj because the bonding is held together by 87 kj so removing that should prevent the bonding from taking place or reverse it I believe. In other words, a certain amount of energy is required to hold the bond together and in the absence of that energy, the bonding will not take place.

7 0

3 years ago