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3 years ago

Why is balance important in organisms?

Chemistry

1 answer:

valentinak56 [21]3 years ago

6 0

Therefore is very important because it ensures survival, existence and stability of the environment.

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What is the main function of the cell membrane in an animal cell?

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Answer:

A cell's plasma membrane, also known as the cell membrane, provides protection. It also maintains a constant atmosphere within the cell, and the membrane serves a variety of purposes. The first is to transport nutrients into the cell, and the second is to transport toxic compounds out

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2 years ago

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What is the difference between qualitative data and quantitative data?

astraxan [27]

Quantitative data can be counted, measured, and expressed using numbers. Qualitative data is descriptive and conceptual.

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3 years ago

True or False: Rubbing a balloon on your head for a longer amount of time will increase the attractive force of the balloon.

Andreyy89

Answer:

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3 years ago

Melatonin has the chemical formula C13H16N2O2. Less than 10 milligrams(0.01g) Convert the melatonin to moles

oee [108]

Answer: There are 0.000043 moles of melatonin.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

given mass = 10 mg = 0.01 g

molar mass of $C_{13}H_{16}N_2O_2$ = 232.28 g

${\text {moles of melatonin}}=\frac{0.01g}{232.28g/mol}=0.000043moles$

Thus there are 0.000043 moles of melatonin

6 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago