CHEMISTRY COLLEGE

To test Döbereiner’s idea, predict:(a) The boiling point of HBr from the boiling points of HCl (- 84.9°C) and HI (-35.4°C) (actual value = -67.0°C)
(b) The boiling point of AsH₃ from the boiling points of PH₃ (- 87.4°C) and SbH₃ (-17.1°C) (actual value = -55°C)

Answers

Answer 1
Answer:

Answer:

a) Approximate boiling point of HBr = -60.15 °C

b) Approximate boiling point of AsH₃ = -52.25 °C

Explanation:

Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)

a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore

Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C

b) Simmilarly for  AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be

Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C

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Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equation.

Answers

Answer:

30 cm³

Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
  • Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

Calculate the mass of a sample of lead (cPb = 0.16 J/g℃) when it loses 200 J cooling from 75.0℃ to 42.0℃.

Answers

looses 299 please collins 75.0

Suppose one was doing this experiment in the real world and the stockroom ran out of NaCl solutions. Which other solutions of similar concentration could be used in place of NaCl(aq)

Answers

Answer:

Polyethylene glycol

Explanation:

Because of the corrosive nature of Brine solution (solution of NaCl) , we can use organic liquid polyethylene glycol .  

What is the molecular formula of each of the following compounds? Part A Empirical formula CHO2, molar mass = 180.0 g/mol

Answers

Answer:

C4H4O8

Explanation:

The molecular formula of a compound with empirical formula: CHO2 can be calculated as follows:

(CHO2)n = 180g/mol

{12 + 1 + 16(2)}n = 180

(12+1+32)n = 180

(45)n = 180

n = 180/45

n = 4.

The molecular formula = (CHO2)4

= C4H4O8

The higher the pH, the less acidic the solution

Answers

If the pH is higher the concentration of hydrogen ions becomes less and the solution becomes less acidic.

As the pH becomes lower, the concentration of hydrogen ions becomes greater, and the solution becomes more acidic.

HOPEIT'SHELPFUL:)

Answer:

yh thats true lol, ty for that very interesting fact

If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?

Answers

Answer:

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Explanation:

Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O

By stoichiometry,

1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O

Therefore, calculating the mass concentration of Fe(NO3)3;

Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))

= 242 g/mol

Mass concentration of Fe(NO3)3 = molar mass * molar concentration

= 242 * 0.2

= 48.4 g/L

Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)

= 242 + 162 g/mol

= 404g/mol

Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass

= 48.4 /404

= 0.12 mol/l

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
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