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Harrizon [31]
2 years ago
5

t physiological pH, the carboxylic acid group of an amino acid will be ________, while the amino group will be ________, yieldin

g the zwitterion form.
Chemistry
1 answer:
astraxan [27]2 years ago
5 0

Answer:

At physiological pH,the carboxylic acid group of an amino acid will be deprotonated while the amino group will be protonated,yielding the zwitter ion form.

Explanation:

Deprotonated means removal of protons in an acid base reaction and protonated means addition of protons in an acid base reaction.

Both protonated and de protonated reaction takes place in catalytic acid bade reaction by changing either it's mass or it's charge.

During formation of zwitter ion, the carboxylic acid will be deprotonated by donating the H+ ion while the amino acid is protonated by taking the H+ ion.

R-CH-COOH                               R-CH-COO-

   I                                ⇒                 I

  NH₂                                              NH₃               (Zwitter ion)

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A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?
kari74 [83]
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon, 
then, (415.2/89.2) ×100 =  465.47 g of the substance will be required to yield 34.6 moles of carbon.
8 0
2 years ago
Read 2 more answers
Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp
weeeeeb [17]

Answer : The grams of N_2 consumed is, 89.6 grams.

Solution : Given,

Mass of CaCN_2 = 265 g

Molar mass of CaCN_2 = 80 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate the moles of CaCN_2.

\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles

The given balanced reaction is,

CaC_2+N_2\rightarrow CaCN_2+C

from the reaction, we conclude that

As, 1 mole of CaCN_2 produces from 1 mole of N_2

So, 3.2 moles of CaCN_2 produces from 3.2 moles of N_2

Now we have to calculate the mass of N_2

\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2

\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g

Therefore, the grams of N_2 consumed is, 89.6 grams.

5 0
2 years ago
Read 2 more answers
3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o
velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

6 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

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

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
Vesnalui [34]
 <span>2 KClO3(s) → 3 O2(g) + 2 KCl(s) 

</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates. 
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 =  3.45 / 122.55 = 0.028

Moles of O2 produce = \frac{3}{2} \times 0.028

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042  = 1.35 g



5 0
2 years ago
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