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Jet001 [13]
3 years ago
8

How many moles are equal to 83.4 L of O2?

Chemistry
1 answer:
ladessa [460]3 years ago
8 0

Answer:

3.72mol

Explanation:

Hello,

In this case, we consider that at STP conditions (273 K and 1 atm) we know that the volume of 1 mole of a gas is 22.4 L, thereby, for 83.4 L, the resulting moles are:

22.4L\rightarrow 1mol\\83.4L\rightarrow X\\X=\frac{83.4L*1mol}{22.4L}=3.72mol

This is a case in which we apply the Avogadro's law which relates the volume and the moles as a directly proportional relationship.

Best regards.

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Explain, in your own words, what volumic mass is
ira [324]

Answer:

The density (more precisely, the volumetric mass density; also known as specific mass), of a substance is its mass per unit volume. The symbol most often used for density is ρ (the lower case Greek letter rho), although the Latin letter D can also be used.

5 0
2 years ago
What is the rate constant of a first-order reaction that takes 5.50 minutes for the reactant concentration to drop to half of it
bulgar [2K]
For a first order reaction, the half life is inversely proportional to the rate constant. 
The formula is
half life = ln(2)/k = 0.693/k
where k is the rate constant

t = 5.50 minutes

k = ln(2)/5.50 = 0.126 min^-1

Your rate constant is 0.126 min^-1.

8 0
3 years ago
Question 3. A batch chemical reactor achieves a reduction in
kotykmax [81]

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

r = k [A]^{x} [B]^{y}

where:

  • [A] is the concentration of species A,
  • x is the order with respect to species A.
  • [B] is the concentration of species B,
  • y is the order with respect to species B
  • k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

v(t) = -\frac{d[A]}{dt} = k [A]^{n}

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

<em>Rate Law:                                    rate = k</em>

<em>Concentration-time Equation:   [A]=[A]o - kt</em>

where

  • k: rate constant [M/s]
  • [A]: concentration in the time <em>t</em> [M]
  • [A]o: initial concentration [M]
  • t: elapsed reaction time [s]

For first-order kinetics, we have:

<em>Rate Law:                                        rate= k[A]</em>

<em>Concentration -Time Equation:      ln[A]=ln[A]o - kt</em>

where:

  • K: rate constant [1/s]
  • ln[A]: natural logarithm of the concentration in the time <em>t </em>[M]
  • ln[A]o: natural logarithm of the initial concentration [M]
  • t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

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The textbook Would most likely have more gravitational potential energy because it is heavier. Things that are heavier have a larger gravitational pull and are pulled to the earth faster
4 0
3 years ago
Oxidation number is
zavuch27 [327]
The correct letter to your answer would be A
5 0
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