All categories

1 year ago

What biological molecules transport other substances, provide structural

Chemistry

1 answer:

PSYCHO15rus [73]1 year ago

3 0

The biological molecules that transport other substances are the proteins.

<h3>What are proteins?</h3>

Proteins are biological macromolecules that contains nitrogen in their molecules and are the building block of a living organism.

The proteins perform the following functions:

they transport other substances,

provide structural support,

cause the movement of muscles, and

catalyze chemical reactions in living organisms.

Learn more about protein here:

brainly.com/question/10058019

#SPJ1

You might be interested in

Is na20 a stable compound, or an unstable compound?

maria [59]

Answer:

unstable

Explanation:

3 0

2 years ago

5 points<br> What is the name of the atom pictured here? Use a periodic table if<br> needed. *

photoshop1234 [79]

Answer:

There is nothing attached to this question so I unfortunately cannot help you

Explanation:

3 0

3 years ago

The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +

3241004551 [841]

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are : a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

3 0

2 years ago

At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil

sertanlavr [38]

Answer: Thus the solubility of $N_2$ gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of $N_{2}$ in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law, $C=k_{H}P$ .

$k_{H}$ = Henry’s law constant = ?

The partial pressure (P) of $N_{2}$ in water = 4.07 atm

\ $C= k_{H}\times P\\95.7mg=k_{H}\times 4.07$

$k_{H}=23.5$

At pressure of 10.0 atm

$C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg$

Thus the solubility of $N_2$ gas in water, at the same temperature, is 235mg/100g

6 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago