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4 years ago

Can someone answer theses problems? Only do the Even number problems,Im in a rush so please help me.

Chemistry

1 answer:

8090 [49]4 years ago

7 0

The answer to the questions. All 15 of them.

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Is it possible to find density and volume of a substance with just the mass?

tensa zangetsu [6.8K]

Even though an atom may be smaller than another atom, it might have more mass. The mass of atoms, their size, and how they are arranged determine the density of a substance. Density equals the mass of the object divided by its volume; D = m/v. Objects with the same mass but different volume have different densities.

7 0

3 years ago

The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.

atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M

Explanation:

The photodissociation of ozone by UV light is given by:

O₃ + hν → O₂ + O (1)

The first-order reaction of the equation (1) is:

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

where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration

We can get the following expression of the first-order integrated law of the reaction (1), by resolving the equation (2):

$[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt}$ (3)

where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration

We can calculate the initial ozone concentration using equation (3):

$[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M$

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!

3 0

4 years ago

Does the entropy (△S) increase or decrease in each of the following processes? The wind blows a pile of leaves all over the yard

garik1379 [7]

B, increasing decreasing is my opinion

6 0

4 years ago

How many square centimeters are in an area of 9.90 inch2 ?

USPshnik [31]

Answer:

63.8708 hope this helps :)

Explanation:

7 0

3 years ago

A balloon inflated in a room at 27 degrees celsius has a volume of 8.00 L.the balloon is then heated to a temperature of 78 degr

dangina [55]

This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K

V₂ = 9.38 L

5 0

3 years ago