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Tems11 [23]
3 years ago
11

What is the original source of energy that is stored in the food that animals eat?

Chemistry
1 answer:
erastova [34]3 years ago
5 0
Since plants turn sunlight into glucose (sugar) that humans and animals eat... the energy we get from our food comes indirectly from the sunlight.
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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]

4 0
3 years ago
I have a balloon that can hold 125,000 mL of air. If I blow up this balloon with 3 moles of oxygen gas at a
matrenka [14]

Answer:

234.35 °C

Explanation:

Given data:

Volume of balloon = 125000 mL

Moles of oxygen = 3 mol

Pressure = 1 atm

Temperature = ?

Solution:

Formula:

PV = nRT

P = Pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

Volume of balloon = 125000 mL × 1 L /1000 mL

Volume of balloon = 125 L

Now we will put the values:

Ideal gas constant = R = 0.0821 atm.L/mol.K

PV = nRT

T = PV/nR

T = 1 atm × 125 L/  0.0821 atm.L/mol.K × 3 mol

T= 125  /0.2463 /K

T = 507.5 K

K to °C

507.5 K - 273.15 = 234.35 °C

4 0
3 years ago
View Lab
Vika [28.1K]

Answer:

c

Explanation:

c

7 0
2 years ago
Read 2 more answers
According to Charles's law, what will happen to the volume of these balloons as temperature
german

Answer:

The volume will decrease and the balloons will be smaller

7 0
2 years ago
Determine the pH of a 0.048 M hypochlorous acid (HClO) solution. Hypochlorous acid is a weak acid (Ka = 4.0 ✕ 10−8 M).
storchak [24]

pH of 0.048 M HClO is 4.35.

<u>Explanation:</u>

HClO is a weak acid and it is dissociated as,

HClO ⇄ H⁺ + ClO⁻

We can write the equilibrium expression as,

Ka = $\frac{[H^{+}] [ClO^{-}]  }{[HClO]}

Ka = 4.0 × 10⁻⁸ M

4.0 × 10⁻⁸ M = $\frac{x \times x }{0.048}

Now we can find x by rewriting the equation as,

x² =  4.0 × 10⁻⁸ × 0.048

   = 1.92 × 10⁻⁹

Taking sqrt on both sides, we will get,

x = [H⁺] = 4.38 × 10⁻⁵

pH = -log₁₀[H⁺]

     = - log₁₀[ 4.38 × 10⁻⁵]

   = 4.35

8 0
2 years ago
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