Cardiovascular and circulatory
kidneys filter thru blood to take out waste
lungs breathe in oxygen, give blood 2 circulatory to carry! takes co2 out
Answer:
H2 > N2 > Ar > CO2
Explanation:
Graham's law explains why some gases efuse faster than others. This is due to the difference i their molar mass. Generally; The rate of effusion of gaseous substances is inversely proportional to the square rot of its molar mass.
This means gases with low molar masses would have higher efusion rate compared to gases with higher molar masses.
So now we just need to compare the molar masses of the various gases;
Ar - 39.95
CO2 - 44.01
H2 - 2
N2 - 28.01
To obtain the order in increasing rate, we have to order the gases in decreasing molar mass. This order of increasing rate is given as;
H2 > N2 > Ar > CO2
The rate law depicts the effect of concentration on reaction rate. Second mechanism 2NO(g) ⇄ N₂O₂(g) [fast], N₂O₂(g) + O₂(g) → 2NO₂(g) [slow] is most reasonable. Thus, option b is correct.
<h3>What is rate law?</h3>
Rate law and equation give the rate at which the reaction takes place under the influence of the concentration of the reactants. The balanced chemical reaction is given as,
2NO(g) + O₂(g) → 2NO₂(g)
The rate of the equation is given as,
rate = k [NO]² [O₂]
In a multi-step chemical reaction, the slowest step is the rate-determining step. The second mechanism is given as,
2NO (g) → N₂O₂ (g) [fast]
N₂O₂(g) +O₂(g) → 2NO₂ (g) [slow]
Rate is given as,
rate = k [N₂O₂] [O₂]
Therefore, option b. the second mechanism is the most reasonable.
Learn more about rate law, here:
brainly.com/question/14779101
#SPJ4
<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
PH scale is used to determine how acidic or basic a solution is.
pH can be calculated as follows;
by knowing the ph we can calculate pOH
pH + pOH = 14
pOH = 14 - 8.1
pOH = 5.9
pOH is used to calculate the hydroxide ion concentration
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 1.26 x 10⁻⁶ M
therefore hydroxide ion concentration is 1.26 x 10⁻⁶ M
