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dedylja [7]
2 years ago
13

Determine which intermolecular forces are the dominant (strongest) forces for a pure sample of each of the following molecules b

y placing the molecules into the correct bins.
Dispersion Forces; Dipole-Dipole Forces; Hydrogen Bonding Forces
Kr, H2O, CHCI3, HF, C2H6, HBr
Chemistry
1 answer:
BartSMP [9]2 years ago
8 0

Answer:

Kr- Dispersion Forces

H2O- Hydrogen Bonding

CHCI3- Dipole-Dipole Forces

HF- Hydrogen Bonding

C2H6- Dispersion Forces

HBr- Hydrogen Bonding Forces

Explanation:

Dispersion forces occurs in all substances. They are the dominant intermolecular interaction in all non polar substances such as C2H6 and Kr.

Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom such as Cl, Br, O etc. It is the dominant intermolecular interaction in HF, HBr and H2O.

Dipole-Dipole interactions occur when a permanent dipole exists in a molecule such as in CHCI3

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TEA [102]

Answer:

Explanation:

Q1.

(a) 46 200

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(b) Energy is used to heat the kettle.

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Q2.

(a) (approximate same size particles as each other and as liquid and gas) touching

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(g) 2 × 4 200 × 801

672 000 (J)

an answer of 672 000 (J) scores 2 marks

[11]

Q3.

(a) x-axis labelled and suitable scale

Page 12 of 13

points plotted correctly

allow 5 correctly plotted for 2 marks, 3−4

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allow ± ½ square

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(b)

allow ecf from line of best fit in part (a)

0.075 (°C/s)

an answer of 0.075 (°C/s) scores 2 marks

(c) Δθ = 11.5 (°C)

a calculation using an incorrect temperature

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ΔE = 1.50 × 900 × 11.5

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an answer of 15.525 (kJ) or 15.53 (kJ) or 15.5

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an answer of 15 525 (kJ) scores 3 marks

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Q4.

(a) 80 °C

ΔE = 0.5 × 3400 × 80

ΔE = 136 000 (J)

an answer of 136 000 (J) scores 3 marks

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allow any sensible practical suggestion

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Page 13 of 13

(d) efficiency = 300/500

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an answer of 0.6 or 60% scores 2 marks

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7 0
2 years ago
A student heats the same amount of two different liquids over Bunsen burners. Each liquid is at room temperature when the studen
ZanzabumX [31]

If Liquid 1 has a higher specific heat than Liquid 2, then Liquid 1 will take longer to increase in temperature because the higher specific heat of a liquid needs more thermal energy for heating a liquid.

<h3>What is specific heat?</h3>

Specific heat of a substance refers to the quantity of heat that is required to raise the temperature of one gram of a substance by one Celsius degree so we can conclude that  Liquid 1 will take longer to increase in temperature

Learn more about heat here: brainly.com/question/24390373

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2 years ago
Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction
Mariana [72]

Answer:

The balanced molecular equation  for the reaction :

Cu(NO_3)_2(aq)+K_2CO_3(aq)\rightarrow CuCO_3(s)+2KNO_3(aq)

Explanation:

The reaction between copper(II) nitrate  and potassium carbonate gives solid precipitate of copper(II) carbonate and aqueous solution of potassium nitrate.

Cu(NO_3)_2(aq)+K_2CO_3(aq)\rightarrow CuCO_3(s)+2KNO_3(aq)

According to reaction, 1 mole of copper(II) nitrate reacts with 1 mole of potassium carbonate to give 1 mole of copper(II) carbonate and 2 moles of potassium nitrate,

8 0
3 years ago
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Citrus2011 [14]

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A volcanic eruption occurs when molten rock, ash and steam pour through a vent in the earth's crust. Volcanoes are described as active (in eruption), dormant (not erupting at the present time), or extinct (having ceased eruption; no longer active).

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4 0
3 years ago
If a volume of gas 177.6mL was collected at a temperature of 25.8C and the pressure is 799.7 torr, what is the original concentr
Step2247 [10]

Answer:

The original concentration of the acid was 0.605 M

Explanation:

Step 1: Data given

Volume of gas = 177.6 mL = 0.1176 L

Temperature = 25.8 °C = 298.95 K

Pressure = 799.7 torr = 799.7/ 760 = 1.0522368 atm

Volume of acid needed to react = 12.6 mL = 0.0126 L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure of the gas = 799.7 torr = 1.0522368 atm

⇒with V = the volume of the gas = 177.6 mL = 0.1776 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 25.8 °C = 298.95 K

n = 0.007618 moles

Step 3: Calculate original concentration

We need 0.007618 moles of acid to react with the same amount of moles gas

Concentration acid = moles / volume

Concentration acid = 0.007618 moles / 0.0126 L

Concentration acid = 0.605 M

The original concentration of the acid was 0.605 M

5 0
2 years ago
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