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nadya68 [22]
4 years ago
15

Wild animals are not considered a natural resource.

Chemistry
1 answer:
Anvisha [2.4K]4 years ago
4 0

Answer:

False

Explanation:

Wild animals are found in an ecosystem and they are considered to be a natural resource and a biotic factor in the environment. They are living things so they are a natural resource.

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Hydrofluoric acid is what type of acid?
posledela

Hydrofluoric acid is a solution of hydrogen fluoride (HF) in water. Solutions of HF are colourless, acidic and highly corrosive. It is used to make most fluorine-containing compounds; examples include the commonly used pharmaceutical antidepressant medication fluoxetine (Prozac) and the material PTFE (Teflon).

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3 years ago
A student has to create a model of a convex lens. Which property of a convex lens should the student include in the model?
Zepler [3.9K]

Answer:

A.  The lens spreads light.

Explanation:

The convex lens will converge the light that passes through it. The direction of light will be bend to be closer to the middle point. This is the lens used for the eyeglass that treat people with hypermetropia. That is why the convex lens also called a converging lens or a plus lens. The strength of the convex lens depends on its focal length, index of the material used, and the radius of curvature.

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3 years ago
Name the following alkenes using systematic names.
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3 0
4 years ago
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Nataly [62]

Answer:

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7 0
3 years ago
Read 2 more answers
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
4 years ago
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