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makvit [3.9K]
3 years ago
11

Relatively small organic molecules with a central carbon atom which is bonded to a carboxyl group, an amino group, a carbon cont

aining group, and a hydrogen atom are called:
Chemistry
1 answer:
Tema [17]3 years ago
8 0
<span>The answer is amino acids. This one is organic compounds containing amine and carboxyl functional group and with a side chain. The elements of this amino are carbon, hydrogen, oxygen and nitrogen and the elements found in the side chains of certain amino acids. It consists of carboxylic acid group and amine. This is called 2-alpha.</span>
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How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the
devlian [24]

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

4 0
2 years ago
Read 2 more answers
Cryogenics has the potential to be useful in a variety of fields, including medicine. Suppose you have engineered a method to su
tatiyna

Explanation:

It is known that the specific heat capacity of Liver (C_{p}) is 3.59 kJ kg^{-1}.K^{-1}

It is given that :

Initial temperature of Liver = Body temperature = 37^{o}C = 310 K

Final temperature of Liver = 180 K

Relation between heat energy, mass, and change in temperature is as follows.

                        Q = m \times C_{p} \times \Delta T

Now, putting the given values into the above formula as follows.

                    Q = m \times C_{p} \times \Delta T

                    Q = 1.5 kg \times 3.59 kJ/kg.K \times (310 - 180) K

                         =  700.05 kJ

Therefore, we can conclude that amount of heat which must be removed from the liver is 700.05 kJ.

7 0
3 years ago
What is the mass of 14.0 mol of Ba(NO3)2?
sashaice [31]

Answer:

261.337 g/mol

Explanation:

Please mark brainliest my answer got taken down its correct

3 0
3 years ago
A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside
Inessa05 [86]
<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

  • Initial volume, V1 = 100.0 mL
  • Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

  • Initial pressure, P1 = 1.80 atm
  • Final temperature , T2 = -25°C

                                     = 248.15 K

  • Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

  • According to the combined gas law equation;

\frac{P1V1}{T1}=\frac{P2V2}{T2}

Rearranging the formula;

V2=\frac{P1V1T2}{T1P2}

Therefore;

V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}

V2=112.08mL

Therefore, the new volume of the gas is 112.08 mL

8 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:
slamgirl [31]

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

8 0
3 years ago
Read 2 more answers
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