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12345 [234]
3 years ago
5

Where in the lungs does gas exchange occur

Chemistry
1 answer:
Lina20 [59]3 years ago
7 0

Answer: Between the alveoli and a network of tiny blood vessels called capillaries, which are located in the walls of the alveoli.

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A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to
Simora [160]

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

8 0
3 years ago
Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) +
VMariaS [17]

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

8 0
3 years ago
How to get left ventricular end diastolic pressure echo equation?
aliina [53]
Whdhrntggmtkasbudubbffjtt
5 0
3 years ago
What pattern appears in the vertical arrangement (columns) of the elements in the Periodic Table, moving from top to bottom?
Grace [21]
Idk if this is the information you need but the energy level gets higher the farther you move down the Periodic Table. Every element in a vertical line has the same energy level.
3 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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