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Arte-miy333 [17]
3 years ago
14

nitrogen is made up of two isotopes N-14 and N15. given nitrogen’s atomic weight of 14.007 what is the percent abundance of each

isotope
Chemistry
1 answer:
Oduvanchick [21]3 years ago
8 0
The answer is n14 if it
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Attempt 4 A galvanic cell has an X electrode with X 2 plus ions in the left beaker and a Y electrode with Y 2 plus ions in the r
babymother [125]

Answer:

See explaination

Explanation:

Since X is more reactive than Y

=> X is oxidized to X2+ and Y2+ is reduced to Y

Overall cell reaction is:

X(s) + Y2+(aq) => X2+(aq) + Y(s)

please kindly see attachment for further solution.

4 0
3 years ago
Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine
andreev551 [17]
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
3 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
What tittle might the indigenous communities of the Americans give the age of exploration
Rom4ik [11]

Answer:

Poop

Explanation:everyone like poop

4 0
3 years ago
A scientist is raising the temperature of a
Arlecino [84]

Answer:

nothing

Explanation:

Here no result have been found yet..

Therfore we cannot determine which variable is dependent and which is not

8 0
1 year ago
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