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kvv77 [185]
3 years ago
14

Is the final answer oxidation or reduction?

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
4 0

Answer:

oxidation

Explanation:

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Can yall plz help me it’s a huge major grade and i don’t know how to do this
fomenos

Answer:

1. 48 mols

2. 0.2 M

5. 1.25 L

Explanation:

Molarity= mols divided by liters

Hope this helps not sure about 3 and 4

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What can we do to protect our oceans ?
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Pick up plastic, reduce waste, reduce pollutants
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You've made the hypothesis that the less rainfall a cactus plant gets, the taller it will grow, since cactuses like dry conditio
Leviafan [203]

A. the height of the cactus plants

Explanation:

The dependent variable in this experiment designed to test this hypothesis is the height of the cactus plants.

In a hypothesis statement, we can always deduce the dependent and independent variables.

  • Independent variables do not rely on other variables. They are usually the cause of the phenomenon observed in an experiment. In this experiment, it is the rainfall on the cactus plant.
  • Dependent variable is that variable that relies on the independent variable. It is usually the effect of changes in independent variable.
  • The height of the cactus plant depends on the amount of rainfall in an area.

learn more:

Controlled experiment brainly.com/question/1621519

#learnwithBrainly

5 0
3 years ago
Which reaction leads to the formation of a precipitate?
Brrunno [24]

2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs

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3 years ago
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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
3 years ago
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