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3 years ago

Why does the moon appear to shine?

Chemistry

2 answers:

lesantik [10]3 years ago

6 0

Answer:

Because it reflects sunlight

Explanation:

go ogle says "The Moon gets its light from the Sun. In the same way that the Sun illuminates Earth, the Moon reflects the Sun's light, making it appear bright in our sky."

olga nikolaevna [1]3 years ago

4 0

it reflects sunlight from the sun

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Please help, will give the brainiest answer

iren2701 [21]

Answer:

Ag+Nano3 yes

MN + Cao.?No

8 0

3 years ago

This is the answer fellas

lubasha [3.4K]

Answer:

Thx man!

Explanation:

Brainliest Please!

5 0

3 years ago

(+)-Carvone and (-)-carvone differ in the orientations of the substituents around which of the following carbon atoms?

myrzilka [38]

Answer:

The correct option is: B. Carbon 5 only

Explanation:

Carvone is a naturally-occurring monoterpenoid consisting of a six-membered cyclic ring. The carbon-5 of this cyclic ring of Carvone is chiral, due to which Carvone exhibits enantiomerism.

The two enantiomeric forms of Carvone are: R-(–)-carvone, or L-carvone, and S-(+)-carvone, or D-carvone.

These two enantiomeric forms differ in the orientation of the substituents on the chiral carbon-5.

Therefore, the correct option is B. Carbon 5 only.

5 0

3 years ago

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

4 years ago

Select the correct order of the GC components

s344n2d4d5 [400]

Carrier Gas, Flow Controller, Column, Detector, Recorder

First we have a cylinder containing the carrier gas. From there, the carrier gas goes to the flow controller, which determines how much carrier gas we are entering into the column (it doesn’t let more gas pass through). Then, the carrier gas enters the column, which is the most important part of the device. The sample enters the column from another place: the injector. Then, the sample and the carrier gas go together across the column. The interactions between the sample and the column will determine how fast each sample component goes through the column, and so: which component gets out earlier. So, at the end, you will have isolated each substance. Then, each one passes (alone) through the detector, which measures something about the sample – this information will let you know which substance it is. Finally, the recorder provides you with the information the detector has found. Nowadays, the recorder is a computer. In the “stone age” they just used a rudimentary printer.

8 0

3 years ago