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3 years ago

9

If a shell can hold a maximum of 32 electrons, what is the value of n

1 answer:

just olya [345]3 years ago

5 0

I think the answe would be 16n

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Why would an atom become an ion?

crimeas [40]

Answer:

c

Explanation: correct me if im wrong

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3 years ago

Which of the following is considered a transition metal?

tatyana61 [14]

Sodium is considered to be a transitional metal

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3 years ago

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A wooden block has a volume of 25.3 cm3. Its mass is 17.0g. What is the density of this block?

kolbaska11 [484]

Density = (mass) / (volume)

Density = (17.0 g) / (25.3 cm³) = 0.672 gm per cm³. (rounded)

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3 years ago

What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C

Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature = $-5.1^oC=273+(-5.1)=267.9K$

Now put all the given values in this formula, we get

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

$\log (\frac{P_1}{100})=-0.576$

$\frac{P_1}{100}=0.265$

$P_1=26.5mmHg$

Thus the vapor pressure of $CS_2CS_2$ in mmHg at 26.5 ∘C is 26.5

7 0

3 years ago

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

2 years ago