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r-ruslan [8.4K]
3 years ago
6

What is the cause of desertification

Chemistry
2 answers:
Dominik [7]3 years ago
8 0

Answer:

C overgrazing is the answer

liq [111]3 years ago
8 0

Answer:

yeah he got it right

Explanation:

it's C

You might be interested in
How many moles of KBr will be produced from 10.51 moles of BaBr2?
gregori [183]

Answer:

21.02moles of KBr

Explanation:

Parameters given:

Number of moles BaBr₂ = 10.51moles

Complete reaction equation:

           BaBr₂ + K₂SO₄ → KBr + BaSO₄

Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:

        The balanced equation is:

           BaBr₂ + K₂SO₄ → 2KBr + BaSO₄

From the equation:

     1 mole of BaBr₂ produces 2 moles of KBr

∴   10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr

7 0
3 years ago
Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t
MariettaO [177]

Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

               HA               ⇄        H⁺        +          A⁻

t= 0      0.200 M                     0                     0

t              -x                             x                       x

t= eq      0.200M -x               x                       x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= \frac{ [H^{+}]  [A^{-} ]}{ [HA]}

Ka= \frac{x x}{0.200 M -x}

Ka= \frac{x^{2} }{0.200 M - x}

From the definition of pH, we know that:

pH= - log  [H⁺]

In this case, [H⁺]= x, so:

pH= -log x

3.0= -log x

⇒x = 10⁻³

We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:

Ka= \frac{(10^{-3})^{2}  }{0.200 - (10^{-3}) }= \frac{10^{-6} }{0.199}= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

5 0
4 years ago
Read 2 more answers
Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t
Veseljchak [2.6K]

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

4NH_3+7O_2\rightarrow 4NO_2+6H_2O

The expression for rate of reaction for the reactant :

\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}

\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}

The expression for rate of reaction for the product :

\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}

From this we conclude that, all the options are correct.

3 0
3 years ago
What is the mass number of an isotope with an atomic number of 57 and that has 69 neutrons?
Karolina [17]

Answer:

mass number=126

Explanation:

57+69=126

7 0
2 years ago
Fluorine-21 has a half life of approximately 5 seconds. You have 340 grams, what fraction of the original nuclei would remain af
madreJ [45]
The half life equation is -->P(t) = Pi (0.5) ^ (t/c)
c is equal to the element to reach its half-life (5 seconds)t is equal to the duration of time the element is expose to (20 seconds)Pi is the initial amount (340)0.5 is the base of this exponential function to represent half-life.P(t) is the expression for the function of time

P(20) = 340 (0.5)^20/5P(20) = 340 (0.5) ^4P(20)= 21.25 grams
Fraction = P(t)/Pi              = P(20)/Pi              =21.25/340              =1/16
Therefore, when given 20 seconds, 340 grams of Fluorine-21 will degrade to 21.25 grams OR 1/16 of its original mass.
Hope this method helps! (This is my answer btw, I think you may have accidentally posted twice?)
7 0
3 years ago
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