Is aluminum a liquid at 1000 kelvin
1 answer:
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What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have
or
where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.
Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
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Thus, we have
P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L
Now, applying these converted values into the gas law, we have
Given that the mass of the sample is 0.430 g, we have
Thus, the gas sample has a molar mass of 43.9 g/mol.
where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.
Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
</span>
Thus, we have
P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L
Now, applying these converted values into the gas law, we have
Given that the mass of the sample is 0.430 g, we have
Thus, the gas sample has a molar mass of 43.9 g/mol.
Answer:
The answer to the question is
The rate constant for the reaction is 1.056×10⁻³ M/s
Explanation:
To solve the question, e note that
For a zero order reaction, the rate law is given by
[A] = -k×t + [A]₀
This can be represented by the linear equation y = mx + c
Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀
Therefore the rate constant k which is the gradient is given by
Gradient = where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M
= = -0.001056 M/s = -1.056×10⁻³ M/s
Threfore k = 1.056×10⁻³ M/s