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3 years ago

Which buffer system is found in the human body?

Chemistry

2 answers:

AveGali [126]3 years ago

7 0

Answer:

Bicarbonate buffer system

Explanation:

The human body uses a bicarbonate buffer system to maintain blood pH at a reasonable level. Without it, we'd be dead.

STatiana [176]3 years ago

7 0

Answer: C. bicarbonate buffer system

The bicarbonate buffer system helps to regulate the pH levels of our body.

A healthy pH range is about 7.35 to 7.45.

Hope this helps!

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Atomic number of f^-1

victus00 [196]

The atomic number is 9 Ions have the same atomic number since atomic number signifies number of protons not electrons

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3 years ago

What is the current produced when a 12-Volt battery encounters a resistance of 20 Ohms?

Verdich [7]

Answer:

Explanation:

V = IR

I = V/R

I = (12V)/(20Ω)

I = 0.6 A

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2 years ago

Plz answer thx How many moles are in 2.5 x 10^18 grams of water?

alex41 [277]

Answer:

$1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

Explanation:

Mass of water = $m = 2.5\times 10^{18} g$

Molar mass of water = M = 18 g/mol

Moles = n = $\frac{m}{M}$

$n=\frac{2.5\times 10^{18} g}{18 g/mol}=1.39\times 10^{17} moles$

So, there are $1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

3 0

2 years ago

A 1-L soda bottle contains dissolved carbon dioxide gas. Which bottle would have the highest concentration of dissolved gas?

yanalaym [24]

The sealed bottle would have the highest concentration of Carbon Dioxide because the pressure is highest and there was no time for diffusion.

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2 years ago

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

7 0

2 years ago