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3 years ago

Which layer of soil often contains litter?

Chemistry

1 answer:

Lunna [17]3 years ago

3 0

The O Layer. I believe. Or the A horizon

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Is it wierd if i think my step sis is thick

iren2701 [21]

Answer:

yes it is and sometimes it's not

8 0

2 years ago

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Sulfur dioxide, SO2(g), can react with oxygen to produce sulfur trioxide, SO3(g), by the following reaction

WITCHER [35]

Answer:

The heat produced is -15,1kJ

Explanation:

For the reaction:

2SO₂+O₂ → 2SO₃

The enthalpy of reaction is:

ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂

As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ

ΔHr = -197,8kJ

Using n=PV/RT, the moles of reaction are:

$n = \frac{1,00atm*3,75L}{0,082atmL/molK*298,15K}$ = 0,153 moles of reaction

As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:

0,153mol× $\frac{-197,8kJ}{2mol}$ = -15,1kJ

I hope it helps!

8 0

2 years ago

Which of the following is a true statement about science?

Papessa [141]

D. All of the answers are true

4 0

3 years ago

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A 35 L tank of oxygen is at 315 K with an internal pressure of 190 atmospheres. How

nata0808 [166]

Answer:

600.7 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = Pressure of oxygen, V = Volume of oxygen, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

From the question,

Given: P = 190 atm, V = 35 L, T = 135 K

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

n = (190×35)/(135×0.082)

n = 600.7 moles of xygen

6 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

Answer: The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

2 years ago