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Masja [62]
3 years ago
10

The diagram below represents a portion of a cell membrane. The arrow indicates that the cell membrane is carrying out the proces

s of...
​
Chemistry
1 answer:
Llana [10]3 years ago
8 0

Answer:

I'm feeling nice today so heres the answer

Explanation:

In the portion of the cell membrane shown in the diagram, the arrow indicates the process of active transport.

Explanation:

Active transport is one of the mechanisms of transmembrane transport, which involves the use of energy. The diagram (see image) shows the hydrogen (H⁺) output from the cytoplasm to the extracellular space, through an H⁺ pump —consuming ATP— which represents an active transport process.

The hydrophobic nature of the cell membrane prevents the free passage of hydrosoluble elements or ions, as H⁺, so they require the use of active transport to pass through it.

The other options presented are not correct, because

Respiration is a process that occurs in the mitochondria.

Diffusion is a passive transport process that does not require energy.

Cellular recognition depends on membrane proteins that act as specific receptors.

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How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
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I’m assuming you mean barium nitrite, Ba(NO2)2.

First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.

45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
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Which diagram is correct
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2 years ago
Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.
natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

pH=4.76+log(\frac{0.95}{0.92})

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

pH=4.76+log(\frac{0.99}{0.88})

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

pH=4.76+log(\frac{0.85}{1.02})

pH = 4.76 - 0.079

pH = 4.681

5 0
4 years ago
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