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stira [4]
2 years ago
14

Given: 12 - x = 20 - 5x To Prove: x = 2

Chemistry
1 answer:
Rina8888 [55]2 years ago
4 0

Answer:

12 - x = 20 - 5x

Add 5x on both sides,

12 - x + 5x = 20 - 5x

12 + 4x = 20

Subtract 12 on both sides,

12 - 12 + 4x = 20 - 12

4x = 8

Divide 4 on both sides,

4x / 4 = 8 / 4

x = 2

Hence proved.

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garri49 [273]
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
                             moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
                                                    = 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane. 
5 0
3 years ago
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If an atom contains one electron and one proton,will it carry any charge or not​
sertanlavr [38]

Answer:

no it will have no charge...it would be electrically neutral! because the number of protons and electrons are equal!

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3 years ago
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En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex
Serga [27]

Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) <em>¿Cuál es la masa de agua a temperatura ambiente?</em>

Podemos determinar la masa inicial del agua (m_{o}), medido en gramos, al conocer su densidad (\rho_{w}), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado (V_{o}), medido en mililitros, a partir de la siguiente expresión:

m_{o} =\rho_{w}\cdot V_{o}

Si sabemos que \rho_{w} = 1\,\frac{g}{mL} y V_{o} = 500\,mL, entonces:

m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)

m_{o} = 500\,g

La masa del agua a temperatura ambiente es de 500 gramos.

2) <em>¿Cuál es la masa de agua cuando se congela?</em>

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) <em>¿Cuál es la masa de agua que queda después de la evaporación?</em>

Durante la evaporación una parte del agua es transferida al aire, entonces podemos calcular la masa final (m_{f}), medido en gramos, de la sustancia al multiplicar el volumen final (V_{f}), medido en mililitros, por la densidad del agua (\rho_{w}), medida en gramos por mililitro,. Es decir,

m_{f} =\rho_{w}\cdot V_{f}

Si sabemos que \rho_{w} = 1\,\frac{g}{mL} y V_{f} = 400\,mL, entonces:

m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)

m_{f} = 400\,g

La masa de agua que queda después de la evaporación es de 400 gramos.

4) <em>¿Qué masa de agua se evaporó? </em>

Determinamos que la masa evaporada de agua (m_{v}), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

m_{v} =m_{o}-m_{f}

Si m_{o} = 500\,g y m_{f} = 400\,g, entonces tenemos que:

m_{v} = 500\,g -400\,g

m_{v} = 100\,g

Se ha evaporado 100 gramos de agua.

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A gaseous substance turns directly into a solid. Which term describes this change?
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Answer:

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Explanation:

Sublmation- solid transforming into a gas, skipping the liquid stage.

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Answer:

Bacterial cell, cell with a nucleus, multicellular organism

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