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3 years ago

For an atoms electrons, how many energy sub levels are present in the principal energy level n=4?

Chemistry

1 answer:

kupik [55]3 years ago

7 0

A. 4. There are four sublevels in the n=4 level.

They are the <em>4s, 4p, 4d</em>, and <em>4f</em> sublevels.

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135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5 C to 100.0 C what is the specific heat of this meta

Harman [31]

The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5.

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3 years ago

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o

Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

$Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}$

Now, we are able to solve for m:

$m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} } = 9.70*10^4 g Na=97 Kg Na$ =97 000 g Na

7 0

3 years ago

Read 2 more answers

The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o

meriva

Answer:

$\boxed{-267.5}$

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

$\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}$

The equation for the reaction is

C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9

$\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}$

3 0

3 years ago

600.0 mL of air is at 20.0 Celsius. What is the volume of the same air at 60.0 Celsius

REY [17]

Answer:

0.682L or 682mL

Use Charles Law of V1/T1 = V2/T2

V1 = 0.6L

T1 = 293K

V2=?

T2= 333K

Explanation:

6 0

3 years ago

Suppose that 4.8 L of methane at a pressure

Ghella [55]

Answer:

972.3 Torr

Explanation:

P2=P1V1/V2

You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.

6 0

3 years ago