True but in more depth they both have the some same qualities in function but provide for each other when one makes oxygen, H2O, and energy and cellular respiration makes CO2 and glucose
Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
V(H</span>₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.<span>
V(H</span>₂SO₄<span>) = 0,0421 L = 42,1 mL.</span>
At constant temperature and pressure, increasing the amount of gas will increase the volume.
Answer: 0.422 M⁻¹s⁻¹
Explanation: <u>Reaction</u> <u>Rate</u> is the speed of decomposition of the reactant(s) per unit of time.
A <u>Rate</u> <u>Law</u> relates concentration of reactants, rate reaction and rate constant:
![r=k[A]^{x}[B]^{y}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D)
where
[A] and [B] are reactants concentration
x and y are reaction order, not related to the stoichiometric coefficients
k is rate constant
r is rate
Before calculating rate constant, first we have to determine reaction order.
In this question, the reactio order is 2. So, the rate law for it is
![-\frac{d[A]}{dt} =k[A]^{2}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dk%5BA%5D%5E%7B2%7D)
and the integrated formula is
![\frac{1}{[A]} =\frac{1}{[A]_{0}} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%7D%20%2Bkt)
in which
[A]₀ is initial concentration of reactant
Then, using initial concentration at initial time and final concentration at final time:
k = 0.422
The rate constant for the reaction is 0.422 M⁻¹.s⁻¹
False I think because they don't use balloons to check the weather
