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2 years ago

Five (5) muscular injuries and disease

Chemistry

2 answers:

seropon [69]2 years ago

7 0

Answer:

tendinitis.

carpal tunnel syndrome.

osteoarthritis.

rheumatoid arthritis (RA)

fibromyalgia.

bone fractures.

Brilliant_brown [7]2 years ago

4 0

Me and the whole world

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Scientists dye cells using fluorescent molecules that bind to the cell membrane so that the cells can be seen under a microscope

xenn [34]

B is correct. cause with microscope you look at the surface

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3 years ago

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Solutes lower the freezing point of water by:

il63 [147K]

Water molecules are constantly moving in Ice they move slower and are more tightly packed together and move at a slower pace to lower the freezing point you would have to find a solute that makes it harder for water to form crystals

7 0

3 years ago

The male monster has an HH genotype, and the female has a hh genotype. Will the baby monsters have horns?

viva [34]

Answer:

like the others its C

Explanation:

the dad has horns but the mom does not so the babys could or they could not

8 0

3 years ago

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Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing

UkoKoshka [18]

Answer: 1.2 moles of carbon dioxide is produced for the given value of oxygen.

Explanation:

The chemical reaction for the combustion of acetylene follows the equation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

By stoichiometry of the reaction:

5 moles of oxygen produces 4 moles of carbon dioxide.

So, 1.5 moles of oxygen will produce = $\frac{4}{5}\times 1.5=1.2mol$ of carbon dioxide.

Hence, 1.2 moles of carbon dioxide is produced for the given value of oxygen.

4 0

4 years ago

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximat

trapecia [35]

Answer:

4.76

Explanation:

In this case, we have to start with the buffer system:

$CH_3COOH~->~CH_3COO^-~+~H^+$

We have an acid ( $CH_3COOH$ ) and a base ( $CH_3COO^-$ ). Therefore we can write the henderson-hasselbach reaction:

$pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}$

If we want to calculate the pH, we have to calculate the pKa:

$pH=-Log~Ka=4.76$

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

$[CH_3COO^-]=[CH_3COOH]$

If we divide:

$\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1$

If we do the Log of 1:

$Log~1=~zero$

So:

$pH~=~pKa$

With this in mind, the pH is 4.76.

I hope it helps!

8 0

3 years ago

Five (5) muscular injuries and disease​

Five (5) muscular injuries and disease