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jonny [76]
3 years ago
10

Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and r

ises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains 3.500 g of mercury and has a capillary that is 0.220 mm in diameter.
How far does the mercury rise in the capillary when the temperature changes from 0.0 ∘C to 25.0 ∘C? The density of mercury at these temperatures is 13.596 g/cm3 and 13.534 g/cm3, respectively.
Chemistry
1 answer:
postnew [5]3 years ago
3 0
Ok yeah a ha yeah wow it is 13.543 g/× compare that with e 13.5
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Why is the energy supplied by the cooker greater than that calculated ?
TEA [102]

Answer:

Explanation:

Q1.

(a) 46 200

accept 46 000

allow 1 mark for correct substitution

ie 0.5 × 4200 × 22 provided no subsequent step

2

(b) Energy is used to heat the kettle.

[3]

Q2.

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

do not accept particles that overlap

regular arrangement (filling the square)

(b) condensing

(c) solid

(d) physical

(e) particles have more kinetic energy

particles move faster

(f) mass of the liquid

specific latent heat of evaporation

(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

correctly plotted for 1 mark

allow ± ½ square

2

line of best fit

(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

scores max 3 marks

ΔE = 1.50 × 900 × 11.5

ΔE = 15 525 (J)

ΔE = 15.525 (kJ)

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

(kJ) scores 4 marks

an answer of 15 525 (kJ) scores 3 marks

[10]

Q4.

(a) 80 °C

ΔE = 0.5 × 3400 × 80

ΔE = 136 000 (J)

an answer of 136 000 (J) scores 3 marks

(b) energy is dissipated into the surroundings

allow any correct description of wasted energy

(c) put a lid on the pan

allow any sensible practical suggestion

eg add salt to the water

Page 13 of 13

(d) efficiency = 300/500

efficiency = 0.6

an answer of 0.6 or 60% scores 2 marks

allow efficiency = 60%

an answer of 0.6 with a unit scores 1 mark

an answer of 60 without a unit scores 1 mark

7 0
2 years ago
What mass of carbon dioxide is formed when 1.75 mol of ethane burns completely in oxygen?
dybincka [34]

Answer:

There is a mass of 154 Grams of Carbon Dioxide.

Explanation:

One mole is equal to 6.02 × 10^23 particles.

This means we have 1.05 X 10^24 total particles of Ethane.

Each ethane particle contains 2 carbon atoms.

If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)

Carbon Dioxide has a molar mass of 44.01 g/mol

So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.

Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.

What you should end up with is 154 Grams of Carbon Dioxide.

Hope this helps (And more importantly I hope I didn't make any errors in my math lol)

As a side note this is all assuming that this takes place at STP conditions.

3 0
3 years ago
Which electromagnetic wave has the longest wavelength?
Vsevolod [243]
Radio waves has the longest wavelength
7 0
3 years ago
Read 2 more answers
Consider the volume of gas collected in the trial in Experiment 2 for Ni at
ZanzabumX [31]

Answer:

Explanation:

H20

5 0
2 years ago
Please need this ASAP. Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone,
stiv31 [10]

Answer:

1.4×10⁸ g of CaO

Explanation:

We'll begin by converting 250 tonnes to grams (g). This can be obtained as follow:

1 tonne = 1×10⁶ g

Therefore,

250 tonne = 250 × 1×10⁶

250 tonne = 2.5×10⁸ g

Next, the balanced equation for the reaction.

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Finally, we shall determine the mass of CaO produced by the decomposition of 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃. This can be obtained as follow:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 2.5×10⁸ g of CaCO₃ will decompose to produce =

(2.5×10⁸ × 56)/100 = 1.4×10⁸ g of CaO.

Thus, 1.4×10⁸ g of CaO will be obtained from 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃.

6 0
3 years ago
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