Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
![K=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
Next, we plug in the given concentrations on the data table to obtain:

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Answer:
364 K or 91°C
Explanation:
Applying,
V₁/T₁ = V₂/T₂................ Equation 1
Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.
make T₂ the subject of the equation,
T₂ = V₂T₁/V₁................. Equation 2
From the question,
Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K
Substitute these values into equation 2
T₂ = (500×273)/375
T₂ = 364 K
T₂ = (364-273) °C = 91 °C
Answer:
the mass of magnesium oxide will be less than the orginal peice of magnesium
Explanation:
because the magnesium burns to ashes in the presense of air
Answer:
n= 0.08186
{He}2s^2 2p^6
Explanation:
PV=nRT
n=PV/RT
n= (1.220 atm)(4.3410 L) / (0.0821 atm*L/mol*K)(788.0 K)
n=0.08186
As for the electron configuration:
Ne:
{He} 2s^2 2p^6
or long hang:
1s^2 2s^2 2p^6
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