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3 years ago

How many grams of product are formed? (Remember the law of conservation of mass) A(20) B(17) C(31) D(34)

1 answer:

6 0

Cjjjjjjhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhuiiiiiiiiiiiiiiiiiiiiiiiiiiiiiujbbbbbbbbbbbbbbbbbb (31) i may be wrong

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When a skin diver is on the top of the water at 1.03 atm, her lung volume is 3.62L. How will the volume of her lungs change as s

Anon25 [30]

Answer:

–2.23 L

Explanation:

We'll begin by calculating the final volume. This can be obtained as follow:

Initial pressure (P₁) = 1.03 atm

Initial volume (V₁) = 3.62 L

Final pressure (P₂) = 2.68 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

1.03 × 3.62 = 2.68 × V₂

3.7286 = 2.68 × V₂

Divide both side by 2.68

V₂ = 3.7286 / 2.68

V₂ = 1.39 L

Finally, we shall determine the change in volume. This can be obtained as follow:

Initial volume (V₁) = 3.62 L

Final volume (V₂) = 1.39 L

Change in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 1.39 – 3.62

ΔV = –2.23 L

Thus, the change in the volume of her lung is –2.23 L.

NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!

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2 years ago

Which of the following environmental risks is most likely to lead to digestive health issues?

algol13

Meat contamination because humans do not eat air or fertilizer so meat contamination is the only logical answer.

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3 years ago

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Which class of compounds does ammonia blog

Igoryamba

T'belongs to the hydrides, it's a nitrogen hydride.

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3 years ago

A molecule involves two atoms in a chemical bond that do what share?

Viefleur [7K]

B. Share Electrons.
Hope this helps.

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3 years ago

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

7 0

3 years ago