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hram777 [196]
3 years ago
15

50 points / will mark BRAINLIEST IF ANSWERED quickly!

Chemistry
1 answer:
liberstina [14]3 years ago
4 0

Answer:

1. density-dependent factors - competition, predation, parasitism, and    disease.

density-independent factors- natural disasters, seasonal cycles, unusual weather, and human activity

2.

As populations increase in size, completion increase as well. For example more people more food demands. as populations become larger and more and more crowded people compete for food, water, sunlight, spaces to live and so on.

3.

Example:

A prolonged drought, with its associated crop loss could cause deaths, financial hardship, and emigration to other areas.

4.

both are under pressure to change in ways that reduce competition, may evolve to occupy different niches

5.

parasites take nourishment from a host - weakening it / killing it

6.

favorable conditions: agriculture, medicine, industry, sanitation

7.

It equal age distribution = very slow, steady growth

large % of children and teenagers = very rapid growth

8.

the recourse may be a part of a organism's niche/habitat (ex. panda and bamboo)

9.

Density (d) depends upon the mass (m) and volume (v) of a substance. The mass depends upon the amount of matter in a substance, whereas volume is the amount of space that a substance takes up.

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Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete
neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = -RTInK_(eq)

where:

R= universal gas constant

T= temperature

K_eq= equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = -RTInK_(eq)

where;

[texK_eq[/tex]=\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

H^+_{inside} ⇒ H^+_{outside}

Equilibrum constant for the transport is given as:

K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}

=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}

[H^+]_{cell}= 10⁻⁷⁴

=3.98 * 10⁻⁸M

[H^+]_{stomach lumen} = 10⁻²¹

=7.94 * 10⁻³M

Hence;

K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}

=\frac{3.98*10^{-8}}{7.94*10{-3}}

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = -RTInK_(eq)

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = -(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = -nFE°_{membrane}

ΔG₂ = -(1 mol)(96.5KJ/mol/V)(60*10^{-3})

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we  can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ G_{total} = G_{1}+G_{2}

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0
3 years ago
Fill in the blank. The amount which an object accelerates
KATRIN_1 [288]
I think the answer is force
6 0
3 years ago
The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c
olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

7 0
3 years ago
A student is filtering a mixture of sand and salt water into a beaker. What will be found in the beaker after the filtration is
Sergio [31]
The student would find the water and sand, because salt dissolves in water unless it was ocean water or sea water
3 0
3 years ago
Read 2 more answers
How many moles of CaCI are present in 60.0mL of a 0.10 M CaCI2 solution?
balandron [24]

Answer:

0.006 mol of CaCl2 are present

Explanation:

Use dimensional analysis for this:

60.0 mL CaCl2/1  * 1 L/1000 mL * 0.10 mol/1 L = 0.006 mol

If you were to write this as fractions out on paper, you'll see that all of the units will cancel out and you'll be left with moles of CaCl2.

**REMEMBER** that molarity (M) is talking about how many moles of that substance is present in one Liter of solvent.

Hope this helps :)

4 0
3 years ago
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