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1 year ago

Describe how sodium conducts thermal energy. [3 Marks]

1 answer:

3 0

In metals, some of the electrons (often one per atom) are not stuck to individual atoms but flow freely among the atoms. Of course, that's why metals are such good conductors of electricity. Now if one end of a bar is hot, and the other is cold, the electrons on the hot end have a little more thermal energy- random jiggling- than the ones on the cold end. So as the electrons wander around, they carry energy from the hot end to the cold end, which is another way of saying they conduct heat.

Here, sodium is a metal which possesses an extra (valence) electron carries the heat around its body as it is a free electron, which enables sodium to conduct thermal energy.

Hope this help :)

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A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temp

MrMuchimi

Answer: The temperature rise is $0.53^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)

c = heat capacity of water = $4.18J/g^0C$

Initial temperature = $T_i$

Final temperature = $T_f$

Change in temperature , $\Delta T=T_f-T_i=?$

Putting in the values, we get:

$5280J=2400g\times 4.18J/g^0C\times \Delta T$

$\Delta T=0.53^0C$

Thus the temperature rise is $0.53^0C$

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2 years ago

The standard reduction potentials for two half-cells involving iron are: given below. Fe2+ (aq) + 2e– ® Fe (s) Eο = –0.44 V Fe3+

matrenka [14]

Answer:

Explanation:

Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V

Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V

Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .

So reaction in the combined cell will be

2Fe⁺³ + Fe = 3Fe⁺²

cell potential = .77 - ( - .44 )

= 1.21 V .

6 0

2 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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1 year ago

Assume that the biocapacity of Earth does not change and that human population growth continues to increase. How will people's e

Alexandra [31]

A. The Average Ecological footprint will need to decrease . 6-11-18

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1 year ago

Read 2 more answers

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

1 year ago