Answer:
Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2
When [NO2] = 0.200 M:
Rate = 10 x 0.200^2 = 0.400 M/s
When [NO2] = 0.100 M:
Rate = 10 x 0.100^2 = 0.100 M/s
When [NO2] = 0.050 M:
Rate = 10 x 0.050^2 = 0.0250 M/s
Explanation:
If you used the method of initial rates to obtain the order for no2, predict what reaction rates you would measure in the beginning of the reaction for initial concentrations of 0.200 m, 0.100 m, & 0.050 m no2.
Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2
When [NO2] = 0.200 M:
Rate = 10 x 0.200^2 = 0.400 M/s
When [NO2] = 0.100 M:
Rate = 10 x 0.100^2 = 0.100 M/s
When [NO2] = 0.050 M:
Rate = 10 x 0.050^2 = 0.0250 M/s
U are in my class because the teacher gave the same question I will tell aleveryone that you are cheat in g
Answer:
B. P (Phosphorus)
Explanation:
The element is be phosphorus.
This is true because phosphorus has a larger atomic radius than carbon. If you move than the group, elements gain additional electron shell. That additional electron shell keeps electrons far from the nucleus of the atom. This actually increases the atomic radius. Also, phosphorus is more electronegative than aluminum because phosphorus itself is a non-metal and non-metals are generally electronegative while aluminum is a metal and electropositive.
Phosphorus has a lower ionization energy than argon because the ionization energy across the period (i.e from left to right) increases and across period 3, you will find phosphorus first before argon in the periodic table.
The answer to this would be 22 mL
The equilibrium membrane potential is 41.9 mV.
To calculate the membrane potential, we use the <em>Nernst Equation</em>:
<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}
where
• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions
• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]
• <em>T</em> = the Kelvin temperature
• <em>z</em> = the charge on the ion (+1)
• <em>F </em>= the Faraday constant [96 485 C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]
• [Na]_o = the concentration of Na^(+) outside the cell
• [Na]_i = the concentration of Na^(+) inside the cell
∴ <em>V</em>_Na =
[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV
