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ziro4ka [17]
3 years ago
9

Based on the properties of the compounds in the interactive, predict whether the given compounds behave as electrolytes or as no

nelectrolytes.
1. LioH
2. C4H2O4
3. LiBr
4. HNo3
Chemistry
1 answer:
Vesna [10]3 years ago
5 0

Explanation:

Before proceeding we have to understand what electrolytes and non electrolytes are;

An electrolyte is a substance that produces an electrically conducting solution when dissolved. An electrolyte is a compound that can dissociate into ions.

Non electrolytes: A substance whose molecules in solution do not dissociate to ions and thus do not conduct an electric current

Going through the options;

1. LiOH

This is a compound of hat would dissociate into Li+ and OH-.  This is an electrolyte.

2. C4H2O4

This is an organic compound. Gnerally organic acids are non electrolytes, with the exception og the acids. This is a nonelectrolyte.

3. LiBr

This is an electrolyte because it would dissociate into Li+ and Br-  ions.

4. HNO3

HNO3 is a strong acid. Because it is a strong acid it will dissociate completely into its ions (H+ and NO3-). Therefore we consider HNO3 to be a strong electrolyte.

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Answer:

0.0953M

Explanation:

First, let us calculate the number of mole present in 12.5g of Na2CrO4.

Molar Mass of Na2CrO4 = (23x2) + 52 +(16x4) = 46 + 52 +64 = 162g/mol

Mass of Na2CrO4 = 12.5g

Number of mole =?

Number of mole =Mass /Molar Mass

mole of Na2CrO4 = 12.5/162 = 0.0772mol

Now, we can calculate for the molarity as follows:

Mole = 0.0772mol

Volume = 810mL = 810/1000 = 0.81L

Molarity =?

Molarity = mole /Volume

Molarity = 0.0772mol/0.81L

Molarity = 0.0953M

4 0
3 years ago
A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T
zhuklara [117]

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

  • Adding solute to water causes the depression of the freezing point.

  • We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

  • <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

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3 years ago
Which of the following statements are TRUE for BOTH oxygen and sulfur atoms?
irina1246 [14]

Answer:

b. have the same kind number of complete shells

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A student fires a bow and arrow in gym class and all his arrows land close to each other, but not on the bullseye. This student
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A student fires a bow and arrow in gym class and all his arrows land close to eachother but not on the bullseye. this student could be said to be:

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4 0
3 years ago
Use the rhombus to find the measure of the following angles m
Studentka2010 [4]

DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°. This can be obtained by understanding the properties of a rhombus.

<h3>What are the required values of sides and angles?</h3>

Here in the question it is given that,

  • ABCD is a rhombus
  • DB = 10, BC = 13, ∠WAD = 25°

We have to find the values of DA, BW, WC, ∠BAC, ∠ACD, ∠DAB, ∠ADC, ∠DBC, ∠BWC.

1) Since all the sides of the rhombus are equal, DA = 13

2) Since diagonals are perpendicular bisectors of each other, BW = 5

3) Since formula of diagonal is,

d₁ = √4a² - d₂,

here a = 13, d₂ = 10

d₁ = √4×13² - 10

d₁ = √676 - 10 = √576 = 24

Since diagonals are perpendicular bisectors of each other

WC = 24/2  ⇒ WC = 12

4) Since diagonals are angle bisectors at the corners, ∠BAC = 25°

5) Since diagonals are angle bisectors at the corners,

∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

Since opposite angles of the rhombus are equal,

∠BAD = ∠BCD = 50° ⇒ ∠BCW = ∠DCW = 25°

⇒ ∠ACD = 25°

6) Since diagonals are angle bisectors at the corners,∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

⇒ ∠DAB = 50°

7) Since adjacent angles are supplementary angles,

∠BAD + ∠ADC = 180° ⇒ 50° + ∠ADC = 180° ⇒ ∠ADC = 180° - 50° = 130°

⇒ ∠ADC = 130°

8) Since diagonals are angle bisectors at the corners,

∠DBC = 130°/2

⇒ ∠DBC = 65°

9) Since all the sides make an angle of 90° at the center,

⇒ ∠BWC = 90°

Hence DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°.

Learn more about rhombus here:

brainly.com/question/27870968

#SPJ9

6 0
2 years ago
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