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dezoksy
17 days ago
8

You want to test the effect of pH on the distribution of Artemia in your 35cm long testing chamber. If you measure a pH = 1 at o

ne end of the chamber and a pH = 13 at the other end, and the gradient distribution along the testing chamber is linear, what would you predict the pH to be 10 cm from the most acidic end?
Chemistry
1 answer:
Tems11 [2.4K]17 days ago
7 0

Answer:The pH measured 10 cm from the most acidic end is 3.42.

Explanation:

The pH at one end = 1The pH at the other end = 13

The chamber length = 13 cm

The change in pH concerning the chamber's length from the acidic end is

Thus, the pH at a distance of 10 cm from the most acidic end is 3.42.

x=\frac{13-1}{35 cm}=\frac{12}{35} pH/cm

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The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the
Alekssandra [2787]

Answer:

Mitochondria are plentiful in mammalian cells, with their proportions varying across different tissues, from less than 1% in white blood cells to as high as 35% in heart muscle cells. It is essential to understand that mitochondria are not static structures but instead form a dynamic network that frequently undergoes processes of fission and fusion. In skeletal muscle, they exist as part of a reticular membrane network. The two subpopulations, subsarcolemmal (SS) and intermyofibrillar (IMF) mitochondria, occupy different subcellular regions and exhibit slight differences in their biochemical and functional characteristics tied to their anatomical context. The SS mitochondria are positioned just beneath the sarcolemma, while IMF mitochondria are found closely associated with myofibrils. Their distinct properties likely play a role in their adaptability. SS mitochondria make up about 10-15% of the total mitochondrial volume and are believed to be more adaptable than their IMF counterparts, despite the latter displaying higher levels of protein synthesis, enzyme activity, and respiration (1).

Explanation:

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1 month ago
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7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.
lorasvet [2589]

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2

7 0
1 month ago
For the following equilibrium: A+2B⇋C+3D If the change in concentration for B is 0.44 M, what will be the change in concentratio
Anarel [2637]
0.22 M
5 0
20 days ago
In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What
Anarel [2637]

Answer:

The force is 38503.5N.

Explanation:

From the problem, we determine:

P (pressure) = 5.00 atm.

Next, to find the force in Newtons (N), we must convert 5 atm into N/m², as shown:

1 atm equals 101325 N/m².

So, 5 atm equals 5 x 101325 = 506625 N/m².

A (the piston area) = 0.0760 m².

Pressure signifies force per unit area, mathematically represented as

P = F/A.

From this, we find F = P × A.

F = 506625 × 0.0760.

Therefore, F = 38503.5N.

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6 0
1 month ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
eduard [2566]

Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

5 0
1 month ago
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