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almond37 [142]
3 years ago
11

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol

ubilities. If more than one answer is correct, enter the letters without delimiting characters. 1. lead chloride A. Zn(CN)2 2. zinc carbonate B. Zn3(PO4)2 C. PbSO4 D. Ni(OH)2 Write the expression for Ksp in terms of the solubility, s, for each salt, when dissolved in water. lead chloride zinc carbonate Ksp = Ksp = Note: Multiply out any number and put it first in the Ksp expression. Combine all exponents for s. Submit Answer

Chemistry
1 answer:
diamong [38]3 years ago
4 0

Answer: ksp= 4s³

Explanation:

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At 506 K and 2.05 atm, if 6.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3? (R = 0.0820
Nady [450]

Answer:

  • 90.4 liter

Explanation:

You must convert the 6.00 grams of H₂ into number of moles, and then use the stoichiometry of the reaction to find the number of moles of NH₃, which can be converted into volume using the ideal gas equation.

<u></u>

<u>1. Number of moles of H</u><u>₂</u>

  • number of moles = mass in grams / molar mass
  • molar mass of H₂ = 2.016g/mol
  • number of moles = 6.00grams / 2.016g/mol = 2.97619mol

<u></u>

<u>2. Number of moles of NH</u><u>₃</u>

<u></u>

i) Chemical equation:

  • 3H₂(g) + N₂(g) →  2NH₃(g)

ii) Mole ratio:

  • 3 mol H₂ : 2 mol NH₃

iii) Proportion:

  • x mol  NH₃ / 2.97619mol H₂ = 3 mol NH₃ / 2 mol H₂

  • x = 4.4642857mol NH₃

<u>3. Volume of NH₃</u>

  • pV = nRT
  • V = nRT/p
  • V = 4.4642857mol × 0.08206 atm·liter/(K·mol) × 506K / 2.05atm
  • V = 90.4 liter
8 0
3 years ago
Based on the evidence which statement differentiates wave A from wave D
valkas [14]

Answer:

wait what a and b sorry

Explanation:

hope we can be friends

can i please get brainliest

5 0
4 years ago
Read 2 more answers
A student put a bottle of water and a can of sugared soda in the freezer to chill them quickly. when she took them out, the bott
fiasKO [112]

The soda water didn't freeze because of the depression in freezing point as compared to the water.

<h3>What is Freezing Point Depression ?</h3>

When a solute is added to a pure solvent , then the value of freezing point is reduced.

The decrease in the freezing point is directly proportional to the molality of the solute.

It is given in the question that

A student put a bottle of water and a can of sugared soda in the freezer to chill them quickly.

when she took them out, the bottle of water was frozen but the can of soda was not.

It is because water is the pure solvent , so freezing point is fixed but the soda water has sugar and Carbon Di oxide along with the solvent water , the presence of the sugar and Carbon -di-oxide decreases the freezing point of the soda water and so it doesn't freezes.

To know more about Freezing Point Depression

brainly.com/question/2292439

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6 0
2 years ago
Which activity best demonstrates Ernest Rutherford’s creativity?
Ber [7]

Answer is: B. designing an experiment to test the plum pudding model.

Rutherford demonstrate that J.J Thompson's Plum Pudding model was not accurate, but before testing, he thought that plum pudding model is correct.

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).  

Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.

That is different from Plum Pudding model, because it shows that most of the atom is empty space.

4 0
3 years ago
Read 2 more answers
Se combustionan 2 g de magnesio en Aire, obteniéndose 3,28 g de óxido de magnesio (MgO)¿Qué masa de oxígeno se consumen en la re
konstantin123 [22]

Answer:

1.30 g

Explanation:

La reacción que toma lugar es:

  • 2Mg + O₂ → 2MgO

Primero <u>convertimos 3.28 g de MgO en moles</u>, usando su <em>masa molar</em>:

  • 3.28 g ÷ 40.3 g/mol = 0.081 mol MgO

Después <u>convertimos 0.081 moles de MgO en moles de O₂</u>, usando los <em>coeficientes estequiométricos</em>:

  • 0.081 mol MgO * \frac{1molO_2}{2molMgO} = 0.0407 mol O₂

Finalmente<u> convertimos 0.0407 moles de O₂ en gramos</u>:

  • 0.0407 mol O₂ * 32 g/mol = 1.30 g
5 0
3 years ago
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