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3 years ago

1. Draw the Dot Structure for Na!

Chemistry

1 answer:

stepladder [879]3 years ago

4 0

Answer:

Na ·

Explanation:

Hello!

In this case, since the Lewis dot structure of any element illustrates the number of valence electrons at the outer shell; we can set up the electron configuration of sodium to obtain:

$Na^1^1\rightarrow 1s^2,2s^2,2p^6,3s^1$

We can see there is only one valence electron, for that reason the Lewis dot structure is:

Na ·

Whereas the dot represents the valence electron.

Best regards!

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What is the symbol for an ion with a 3+ charge, 28 electrons, and a mass number of 71

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Answer:

Ga3+ is a gallium cation

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2 years ago

If the atmospheric pressure is 0.975 atm what is the pressure of the enclosed gas

Nuetrik [128]

Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm.
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
= 0.975 - (520/760)
= 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm
So the pressure of the gas =P(atm) + Height (Hg)
= 0.975 + (67/ 760) = 1.06 atm
Picture (3)
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
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2 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

Best regards.

7 0

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3 years ago