Describe what happened on the Passover day at the time of jesus

Chemistry

1 answer:

iren [92.7K]3 years ago

3 0

Answer:

Some Christians observe the celebration on the day before Passover, at the same time that Jesus held his Last Supper, while others observe it at the same time that the Passover was sacrificed, that is, the time of Jesus' death, which occurred "at the ninth hour" of the day

Explanation:

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Answer:

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where R is the molar constant of the gases and n the number of moles.

In this case you know:

P= 7.5 atm
V= 2.3 L
n= ?

R= 0.082 $\frac{atm*L}{mol*K}$

T= 24 °C= 297 °K (being 0°C=273°K)

Replacing:

$7.5 atm* 2.3 L=n*0.082 \frac{atm*L}{mol*K} *297K$

Solving:

$n=\frac{7.5 atm* 2.3 L}{0.082 \frac{atm*L}{mol*K} *297K}$

n=0.708 moles

Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?

$mass=\frac{0.708 moles*32 grams}{1mole}$