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3 years ago

Which statement accurately describes a light-year

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

A light-year is a unit of distance. It is the distance that light can travel in one year. Light moves at a velocity of about 300,000 kilometers (km) each second. So in one year, it can travel about 10 trillion km.

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Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain.

pashok25 [27]

I think the the state <span>of the air in an isolated room is completely specified by the temperature and the pressure. Since it is isolated, the volume would be the same as the room. Gas is specified by temperature, pressure and volume. Having the values for pressure and temperature then the gas is completely specified. Hope this answers the question.</span>

3 0

3 years ago

Read 2 more answers

Please help me with number 5

mrs_skeptik [129]

Fluorine - 18.998 u
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tellurium - 127.6 u
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4 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

stiv31 [10]

Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's y = C

y = 3x

Still not converted sum of reaction

for A: 100 - 2x

for B: 50 - x

Shift of x over time

$\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}$

Integration of x as regards t

$\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\$

D is the constant of integration

initial conditions: t = 0, x = 0

$\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\$

hence we get:

$\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}$

after t = 7 minutes , $C = 10 \ g = 3x$

$3x = 10\\\\x = \frac{10}{3}$

Insert the above value x into $\frac{1}{(100-2x)}$ equation $= kt + \frac{1}{100}$ to get k.

$\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\$

$\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091$

therefore plugging in the equation the above value of k

$\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}$

Let y = C

, calculate C:

y = 3x

$y =3 \times \frac{-45.5t}{1+0.91t}$

amount of C formed in 28 mins

$x = \frac{-45.5t}{1+0.91t} ,$ plug t = 28

$\to x = \frac{-1274}{1+25.48} \\\\\to x = \frac{-1274}{26.48} \\\\\to x= -48.26$

therefore amount of C formed in 28 minutes is = 3x = 144.78 grams

C: $y =3 \times \frac{-45.5t}{1+0.91t}$

y= 136.5 =137

7 0

3 years ago

Plz help plz help me will mark Brainly

strojnjashka [21]

Answer:

Greenhouse gases absorb some of the energy and trap it in the lower atmosphere. Less heat radiates into space, and Earth is warmer. Many greenhouse gases occur naturally. Carbon dioxide, methane, water vapor, and nitrous oxide are naturally present in Earth's atmosphere. Since some of the extra energy from a warmer atmosphere radiates back down to the surface, Earth's surface temperature rises.

Hope this helped! :)

6 0

3 years ago

Based on the six given solvents, which of the following answers lists ALL of the solvents that are NOT suitable for liquid-liqui

Zina [86]

Answer:

I don't know the answer sorry

6 0

3 years ago