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jonny [76]
4 years ago
10

If the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, how high a column of water in meters can be suppor

ted by standard atmospheric pressure?
Chemistry
1 answer:
dexar [7]4 years ago
4 0

Answer:

The answer to this is

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

Explanation:

To solve this we first list out the variables thus

Density of the water = 1.00 g/mL =1000 kg/m³

density of mercury  = 13.6 g/mL = 13600 kg/m³

Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals

Therefore from the equation for denstity we have

Density = mass/volume

Pressure = Force/Area  and for  a column of water, pressure = Density × gravity×height

Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa

This value of pressure should be supported by the column of water as follows

Pressure = 101396.16 Pa =  kg/m³×9.81 m/s² ×h

∴  h = \frac{101396.16}{(1000)(9.81)} = 10.336 meters

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

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Answer:

1-(tert-butoxy)-2-methylpropane

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Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we        start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

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3 0
4 years ago
Consider the following reaction: AB(s)⇌A(g)+B(g) . At equilibrium, a 10.9 L container at 650 K contains A(g) at a pressure of 0.
irina1246 [14]

Answer:

pA = 0.095 atm

pB = 0.303 atm

Explanation:

Step 1: the reaction

AB(s) ⇔ A(g) + B(g)

Kp = pA * pB

⇒ with Kp = equilibrium constant

Kp = 0.126 * 0.23  ⇒ Kp = 0.02898

Since the container will be compressed to half of its original volume, means that he pressure will be doubled.

⇒pA = 0.252

⇒pB =0.46

To establish this equilibrium, each pressure has to be lowered by x

⇒pA = 0.252 - x

⇒pB = 0.46 - x

Kp = 0.02898 = (0.252 - x)(0.46-x)

0.02898 = 0.11592 - 0.252x -0.46x + x²

-x² + 0.712x - 0.08694 = 0

D= b² - 4ac

⇒ D = 0.712² -4*(-1) *(-0.08694) = 0.506944‬ -0.34776‬ =0.159184

x = (-b ± √D)/2a  

x = (-0.712 ± √0.159184)/(2*-1)  = (-0.712 ± 0.398978696)/-2

x = 0.156510652 or x= 0.555489348

x = 0.555489348 is impossble or the pressure would be negative

x=0.156510652

pA =0.252 - 0.156510652 = 0.095489348 atm

pB = 0.46 - 0.156510652 = 0.303489348 atm

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