Why is it that you can not feel earths movement but you can observes its effects

1 answer:

5 0

The correct answer would be we have no other frame of reference besides the sky. We could actually tell we are moving because of the shifting of stars in the sky over time. But really, “A” would be your answer. We have the same momentum of the earth. It’s like being on a train at a constant speed. The only way you know you’re moving is if you look out the window or the speed changes causing you to be pushed around. Same thing with earth, our only reference is the sky

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1. 17.0 grams of xenon hexafluoride is in a solid container. How many milliliters of that gas

BlackZzzverrR [31]

Answer: The volume of gas is 3020 ml

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 821.4 torr = 1.08 atm (760 torr = 1atm)

V = Volume of gas in L = ?

n = number of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{17.0g}{245.28g/mol}=0.069mol$

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $302.7^0C=(302.7+273)K=575.7K$

$V=\frac{nRT}{P}$

$V=\frac{0.069mol\times 0.0821Latm/K mol\times 575.7K}{1.08atm}=3.02L=3020ml$

Thus volume of gas is 3020 ml

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aliya0001 [1]

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Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$