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Cerrena [4.2K]
3 years ago
9

All of the alkali earth metals, Group 2, have two valence electrons. Which of these would represent the oxidation number of the

alkali earth metals such as magnesium and calcium?
A) -6
B) -2
C) +2
D) +6

Chemistry
1 answer:
juin [17]3 years ago
5 0
Your answer should be C.) +2. "All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2." 


Credits: https://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Blo...

Hopefully this has helped! :)
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Osmosis occurs when two solutions of different concentrations are separated by a semipermeable membrane. The membrane allows sol
Finger [1]

Answer:

Part A → 7.82 atm

Part B → The unknown solution had the higher concentration

Part C →  0.83 mol/L

Explanation:

Part A

Osmotic pressure (π) = M . R. T . i

NaCl → Na⁺  +  Cl⁻ (i =2)

0.923 g of NaCl must be dissolved in 100 mL of solution.

0.923 g / 58.45 g/m = 0.016 moles

Molarity is mol/L → 0.016 m / 0.1L = 0.16M

π = 0.16M . 0.08206 L.atm/molK . 298K . 2 ⇒ 7.82atm

Part. B

The solvent moves toward the solution of higher concentration (to dilute it) until the two solutions have the same concentration, or until gravity overtakes the osmotic pressure, Π. If the level of the unknown solution drops when it was connected to solution in part A, we can be sure that had a higher concentration.

Part. C

π = M . R . T

20.1 atm = M . 0.08206 L.atm/mol.K . 294K

20.1 atm / (0.08206 L.atm/mol.K . 294K) = 0.83 mol/L

8 0
3 years ago
One gram of an acid, HA of unknown molecular mass is dissolved in water and 22.72 cm3 of 0.60 M sodium hydroxide solution added
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3 years ago
When a 1.06 g sample of Compound Q, a nonelectrolyte, is dissolved in 11.6 g of water, the boiling point of the resulting soluti
Vesna [10]

Answer:

The molar mass of compound Q is 100 g/mol

Explanation:

Step 1: Data given

Mass of compound Q = 1.06 grams

Mass of water = 11.6 grams

Boiling point of the solution = 100.47 °C

Boiling point of water = 100.00 °C

Kb for water = 0.512 °C/m

Step 2: Calculate molality

ΔT = i*kb*m

⇒ with ΔT = the boiling point elevation = 0.47 °C

⇒ with i = the van't Hoff factor = 1

⇒ with kb = the boiling point elevation constant = 0.512 °C/m

⇒ with m = the molality = moles compound Q / mass water

m = ΔT / (i*Kb)

m = 0.47 / 0.512

m = 0.918 molal

Step 3: Calculate moles of Q

Molality = moles Q / mass H2O

moles Q = 0.918 molal * 0.0116 kg

moles Q = 0.0106 moles

Step 4: Calculate molar mass

Molar mass Q = mass Q / moles Q

Molar mass Q = 1.06 grams / 0.0106 moles

Molar mass Q = 100 g/mol

The molar mass of compound Q is 100 g/mol

3 0
3 years ago
Drag the labels onto the table to indicate when each statement Is true. Labels can be used once, more than once, or not at all.
liq [111]

The labels the table to indicate when each statement Is true. Labels can be used once, more than once, or not at all, The orange dye moves independently of the purple dye. 2. Concentration gradients exist that drive diffusion of both dyes. : <u>free water, solute, free water, solute.</u>

In chemistry, attention is the abundance of a constituent divided by way of the total quantity of an aggregate. several types of mathematical descriptions may be prominent: mass concentration, molar concentration, range concentration, and quantity concentration.

it's miles the amount of solute dissolves in one hundred g solvent. If the attention of the answer is 20 %, we understand that there are 20 g solutes in one hundred g solution. instance: 10 g salt and 70 g water are mixed and the solution is ready. find awareness of the answer by means of percentage mass.

Learn more about concentration here:

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3 0
1 year ago
Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad
ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH    +     OH⁻         ⇄        HCOO⁻     +    H₂O

20 mmol       36 mmol             20 mmol

                    16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO  →  Na⁺  +  HCOO⁻

HCOO⁻  +  H₂O  ⇄   HCOOH  +  OH⁻   Kb →  Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So:  [OH⁻] =  16 mmol / 400 mL →  0.04 M

- log  [OH⁻]  = pOH →  1.39

pH = 14 - pOH → 12.61

6 0
3 years ago
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