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IrinaK [193]
2 years ago
7

You have two containers of equal volume. one is full of helium gas. the other holds an equal mass of nitrogen gas both gases hav

e the same pressure how does thegas. bothgaseshavethesamepressure. howdoesthetemperature of the helium compare to the temperature of the nitrogen?
Chemistry
1 answer:
alexira [117]2 years ago
8 0

Let's assume that both He and N₂ have ideal gas behavior.<span>

Then we can use ideal gas law,
     PV = nRT
Where, P is the pressure of gas, V is the volume, n is moles of gas, R is universal gas constant and T is the temperature in Kelvin.

<span>The </span>P <span>and </span>V <span>are </span>same<span> for the both gases.</span>
R is a constant.

The only variables are n and T.

<span>Let's say temperature of </span>He<span> <span>is </span></span>T</span>₁<span> <span>and temperature of </span></span>N₂<span> <span>is </span></span>T₂.<span>

n = m/M<span> where n is moles, m is mass and M is molar mass.</span>

Molar mass of He is 4 g/mol and molar mass of N₂ is 28 g/mol</span><span>

<span>Since mass (m) of both gases are same,</span>
 moles of He = m/4
 moles of N₂ = m/28</span><span>

Let's apply the ideal gas equation for both gases.
For He gas,
 PV = (m/4)RT₁              </span>(1)<span>

For N</span>₂ gas,<span>
 PV = (m/28)RT₂<span>           </span></span> (2)<span>

(1) = (2)
</span><span>(m/4)RT₁ = (m/28)RT₂</span> <span>
        T₁/4 = T₂/28</span><span>
        T₁    = T₂/7</span><span>
<span>        </span>7T</span>₁  = T₂<span>

Hence, the temperature of N</span>₂<span> gas is higher by 7 times than the temperature of He gas.</span>

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Alla [95]
The title of Scientist was formally bestowed upon Sir Isaac Newton when he was
awarded the Merit Badge in Science at the age of 15, and he remained a Scientist
until he died, at the age of 84, on March 20, 1727, for a total duration of 69 years.

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5 0
2 years ago
Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A
lesantik [10]

<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.

<u>Explanation:</u>

The chemical formula of carbonic acid is H_2CO_3. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

               H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)

The expression of first equilibrium constant equation follows:

Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}

The chemical equation for the second dissociation of carbonic acid follows:

               HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0
2 years ago
Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to
Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode (m_{ch}) is given as :

\frac{m_{ch}}{M} ZF = It

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; (m_{ch}) = (Ad) \rho ; replacing that into above equation; we have:

\frac{(Ad) \rho}{M} ZF = It  ---- equation (1)

where;

A = area

d = thickness

\rho = density

From the above equation (1); The time required for coating can be calculated as;

[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons }  ] = (20 A) t

t = \frac{2100}{20}

= 105 s

8 0
2 years ago
How many moles of carbon are in 7.87x10^7 carbon molecules?
zaharov [31]
There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
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6 0
2 years ago
What is the concentration of Iodine I2 molecules in a solution containing 2.54 g of iodine 250 cm3 of solution? A 0.01mol/dm3 B
gulaghasi [49]

Answer:

C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (c), measured in moles per cubic decimeter, can be determined by the following formula:

c = \frac{m}{M\cdot V} (1)

Where:

m - Mass of iodine, measured in grams.

M - Molar mass of iodine, measured in grams per mol.

V - Volume of solution, measured in cubic decimeters.

If we know that m = 2.54\,g, M = 253.809\,\frac{g}{mol} and V = 0.25\,dm^{3}, then the concentration of iodine in a solution is:

c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}

c = 0.04\,\frac{mol}{dm^{3}}

Hence, the correct answer is C.

3 0
2 years ago
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