Suppose the ideal volume for an airbag is 50.0L. Using the pressure of 765.0 mmHg and temp of 29.5C how many grams of Nan3 shoul
1 answer:
Answer:
130 gram of NaN₃should be put into the compartment
Explanation:
Given that Ideal volume for an airbag = 50 lit
Pressure = 765 mmHg
we know
∴
Temperature (T) = 29.5°C = (29.5 + 273) = 302.5 K
We know A/c to ideal gas equation
PV = nRT
Where n= no. of mole
R = universal gas constant = 0.0821 lit. atm. mol⁻¹K⁻¹
Mass of (NaN₃) = mole x molar mass = 2 x 65 = 130 gram
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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
Answer:
147.2g
Explanation:
The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density
