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3 years ago

How does the concentration of a gas in solution change when the partial pressure of the gas above the solution increases?

1 answer:

5 0

Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures

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The elements found in Group 2A, from top to bottom, include beryllium, magnesium, calcium, strontium, and barium. Which of these

elena-14-01-66 [18.8K]

Electronegativity is the property of the element to attract electrons to its nucleus. The trend of electronegativity in the periodic table is decreasing from right to left and decreasing from top to bottom. hence from the given elements, the element with the highest electronegativity should be A.beryllium

5 0

4 years ago

Read 2 more answers

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

4 years ago

The rate of acceleration of an object is determined by the mass of the object and...

GREYUIT [131]

Answer:

a. forces acting on the object

7 0

3 years ago

To answer this question, you may need access to the periodic table of elements.

OleMash [197]

Answer:

b.) Br and Br

Explanation:

A covalent bond occurs when electrons are shared between two atoms causing them to form a bond.

A "pure" covalent bond refers to a nonpolar covalent bond. In these bonds, the electrons are shared equally between two atoms as a result of the absence of an (or very small) electronegativity difference. The purest covalent bond would therefore be between two atoms of the same electronegativity. Two bromines (Br) have the same electronegativity, thus making it the purest covalent bond.

Polar covalent bonds occur when electrons are shared unequally between two atoms. There is a larger electronegativity difference between the two atoms, but not large enough to classify the bonds as ionic. In this case, a.) and c.) are polar covalent bonds and d.) is an ionic bond.

7 0

1 year ago

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp

Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

Equilibrium when T = 100 °C

Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must add heat to boil water

ΔS > 0 (positive), because changing from a liquid to a gas increases the disorder

ΔG = 0, because the liquid-vapour equilibrium process is at equilibrium at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.

If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.

If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0

3 years ago