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Fofino [41]
2 years ago
11

(07.07)What transformation has changed the parent function f(x) = log2 x to its new appearance shown in the graph below?(6 point

s)
f(x + 3) + 2
f(x − 3) − 2
f(x + 2) + 3
f(x − 2) − 3
Chemistry
1 answer:
amid [387]2 years ago
7 0
The answer is B. f ( x - 3 ) - 2





























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A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
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Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

The volume of the gas after it is heated.

Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L

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The following question appears on a quiz: ""You fill a tank with gas at 60°C to 100 kPa and seal it. You decrease the temperatu
dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

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P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K

Putting values in above equation, we get:

\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa

Hence, the final pressure will decrease ad the value is 85 kPa

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