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3 years ago

What acceleration does the force of earths gravity produce?

Chemistry

2 answers:

valentina_108 [34]3 years ago

8 0

(a) 9.8 m/s/s

This never changes.

SCORPION-xisa [38]3 years ago

7 0

Answer: A

Explanation: Acceleration due to gravity, denoted by (g), is the acceleration gained by an object because of gravitational force action on the object. It has a standard value 9.8m/s/s.

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Why imparities reduces boiling point but increasing melting point

dmitriy555 [2]

Answer:

Adding impurities to a solution, in most cases, increases the boiling point of the solution. This occurs because the presence of impurities decreases the number of water molecules available to become vaporized during boiling.

Explanation:

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4 years ago

How do the circulatory, respiratory, or digestive systems work together?

Alexandra [31]

Answer:

by pumping blood throughout the body.

Explanation:

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3 years ago

What is Le chatelier's principle?

ser-zykov [4K]

Answer:

Le Chatelier's principle, also called Chatelier's principle or "The Equilibrium Law", is a principle of chemistry used to predict the effect of a change in conditions on chemical equilibria.

hope it helps ,pls mark me as brainliest

7 0

3 years ago

Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces

inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>

aX → bY (1)

$rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}$

<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>

(a) Based on the definition above, we can express the rate of reaction (2) as follows:

3O₂(g) → 2O₃(g) (2)

$rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}$ (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:

$rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}$

$+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}$

$\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}$

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

6 0

4 years ago

A factory worker was splashed in the eyes with a strong acid chemical. he complains of intense pain and blurred vision. your amb

mote1985 [20]

Flush eyes with plain water for 20-25 min

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3 years ago