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Paladinen [302]
2 years ago
10

Be sure to answer these questions in your journal entry: • How long was the journey to Earth? • How fast did you travel? • Was y

our speed affected when you entered Earth’s atmosphere? If so, how? • Was your speed affected when you traveled through the glass of the window? If so, how? • What was the journey like to get to the retina? PLZ HELP ME

Chemistry
1 answer:
damaskus [11]2 years ago
3 0

Answer: What you doing

Explanation:

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D will be the answer I hope that help good luck I hope. My answer help you
7 0
3 years ago
Define all the physical features of earth​
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5 0
3 years ago
Read 2 more answers
An 80.0g sample of an unknown metal is at an initial temperature of 55.5oC. Afer 540 J of energy is absorbed by the metal, the t
Lunna [17]

Answer:

Specific heat of metal = 0.26 j/g.°C

Explanation:

Given data:

Mass of sample = 80.0 g

Initial temperature = 55.5 °C

Final temperature = 81.75 °C

Amount of heat absorbed = 540 j

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  81.75 °C - 55.5 °C

ΔT =  26.25 °C

540 j = 80 g × c × 26.25 °C

540 j = 2100 g.°C× c

540 j / 2100 g.°C = c

c = 0.26 j/g.°C

7 0
3 years ago
At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO
erastovalidia [21]

Answer : The concentration of NO at equilibrium is 0.9332 M

Solution :  Given,

Concentration of N_2 and O_2 at equilibrium = 0.200 M

Concentration of N_2 and O_2 at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}

K_c=6.25

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

                         N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially             0.200   0.200         0

.800

At equilibrium  (0.200-x) (0.200-x)  (0.800+2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}

By solving the term x, we get

x=2.6\text{ and }-0.0666

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of NO at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of NO at equilibrium is 0.9332 M

5 0
3 years ago
Which is an example of how society affects science?
Evgesh-ka [11]

Answer: Laws have been passed banning the production of a living copy of a person.

i just took the test so i know the answer is correct.

6 0
3 years ago
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