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4 years ago

Which statements correctly describe atmospheric pressure?

Chemistry

1 answer:

djyliett [7]4 years ago

7 0

Answer: Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.

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8. As the temperature of a mixture increases, one part of the mixture may 2 points

disa [49]

Answer:

It's false

Explanation:

Mixtures are always combinations of the same compounds that are at different states.

8 0

3 years ago

Someone who is good at chemistry please help me

erastovalidia [21]

Equation#1 - Ca(s)+Cl2(g) ——> CaCl2(s)
Word equation #1- Calcium and Chlorine react to form Calcium Chloride.

Equation #2- N + I3 —-> I3N
Word Equation #2- Nitrogen and Iodine react to form nitrogen triiodine.

6 0

3 years ago

4.337 + 84.7123 ? i need help lol

ziro4ka [17]

Answer:

89.0493

Explanation:

Add the numbers(line up the decimals)

4.337 + 84.7123

Than you should get

89.0493

4 0

3 years ago

vovikov84 [41]

The chemical reaction will be NaF

3 0

4 years ago

Read 2 more answers

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, the 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:

aX → bY (1)

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

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

(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