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2 years ago

Warm air rises at the equator and cold air sinks at the poles creating

Chemistry

1 answer:

andrey2020 [161]2 years ago

7 0

Answer:

c.convention currents

Explanation:

As hot air cools it sinks back to the surface of the earth, where it gets warmed by the ocean only to rise again.This is called a convection current.

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If a rock had a mass of 20g explain how you would fund the density

REY [17]

Answer:

you have to divide it

Explanation:

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2 years ago

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Alkanes do not have geometric isomers because the carbon atoms in their carbon-carbon bonds are _____.

Mariana [72]

The word that best fits the underlined in the sentence is "free-to-rotate." The carbon atoms in their carbon bonds are free to rotate since alkanes do not have geometric isomers. They only have single bonds and the most common example of which are trans molecules.

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2 years ago

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a

s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

Explanation:

Partial pressure of the $O_2$ gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

$p_{o_2}=K_H\times\chi_{O_2}$

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

$0.0013=34860 bar\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}$

Let the number of moles of $O_2$ gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L $n_w=\frac{1000 g}{18 g/mol}=55.55 mol$

$\chi_{O_2}=\frac{n}{n+n_w}$

$2.56\times 10^{-5}=\frac{n}{n+55.55}$

$n=1.43\times 10^{-7} moles$

$Molarity=\frac{\text{Moles of}O_2}{Volume}$

Molar concentration of oxygen gas in 1 L of water:

$=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L$

The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

4 0

2 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

2 years ago

sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5. V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a) ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0

2 years ago