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Anastasy [175]
4 years ago
7

Which term describes this reaction? H 3 N superscript plus bonded to C, which is bonded above to R subscript 1, below to H, and

right to a second C, which is bonded below right to O superscript minus and double bonded above right to O; plus H bonded to N superscript plus, which is bonded above and below to N and to the right to C, which is bonded above to R subscript 2, below to H and right to a second C, which is bonded below right to O superscript minus and double bonded above right to O; right arrow, with downward arrow splitting off labeled H 2 O; H 3 N superscript plus bonded to C bonded to R subscript 1 above, H below, and a second c to the right; that second C is double bonded above to O and single bonded t the right to N, which is single bonded below to H and right to C, which is bonded above to R subscript 2 below to H and right to a final C, which is bonded below right to O superscript minus and double bonded above right to O. Addition condensation elimination substitution

Chemistry
2 answers:
QveST [7]4 years ago
3 0

Answer:

Condensation reaction.

Explanation:

The reaction is that of a condensation of two amino acids through peptide linkage to form a dipeptide with the elimination of a water molecule.

Check the attachment below for molecular structures of the reactants and product.

Gwar [14]4 years ago
3 0

Answer:

C

Explanation:

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Please help me! I don't understand this at all. All the info is in the picture. Thank you so much!!!
kykrilka [37]

Answer:

H₂S + Cl₂ —> S + 2HCl

Explanation:

? + Cl₂ —> S + 2HCl

To balance the equation above, we must recognise what atoms are present in the products.

The products contains S, H and Cl.

Thus, S, H and Cl must also be present in the reactants.

Considering the equation given above, we can see clearly that H and S is missing in the reactants.

H and S together as a compound is expressed as H₂S.

Now, we shall input H₂S into the equation to obtain the complete equation. This is illustrated below:

? + Cl₂ —> S + 2HCl

H₂S + Cl₂ —> S + 2HCl

Next, we shall verify to see if the equation is balanced.

There are 2 atoms of H on both sides of the equation.

There are 2 atoms of Cl on both sides of the equation.

1 atom of S exist on both sides of the equation.

Thus, the equation is balanced.

3 0
3 years ago
How many molecules are in 25g of Na2SO4? POINTS!! Please help! Im so frustrated.
erik [133]
Hey there!

Molar mass Na2SO4 = <span>142.04 g/mol

Number of moles:

n = m / mm

n = 25 / 142.04

n = 0.176 moles of Na2SO4

Therefore, </span>use the Avogadro constant


1 mole Na2SO4 ------------------- 6.02x10²³ molecules
0.176 moles Na2SO4 ------------   molecules ??


0.176  x  ( 6.02x10²³ ) / 1 

=> 1.059x10²³ molecules of Na2SO4

hope this helps!
6 0
4 years ago
I need help :(((((((((((((((
GarryVolchara [31]
Can I see the article and what your researching
3 0
3 years ago
If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of      sand, what  percent by mass of
disa [49]
3.84 - 1.43 = 2.41
2.41g of table sugar

% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76

62.76%
8 0
3 years ago
Read 2 more answers
This is due at 2:00 am today. I only have a few points. No links please I need a real answer.
GuDViN [60]

Answer:

7.5 g of AlCl3

Explanation:

The given equation is;

NaOH + AlCl3 --> Al(OH)3 + NaCI.

By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.

Thus, let's make them equal by balancing the equation.

Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:

3NaOH + AlCl3 --> Al(OH)3 + 3NaCI

We can see that;

NaOH has 3 moles

While AlCl3 has 1 mole

Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;

2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3

5 0
3 years ago
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