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Sveta_85 [38]
1 year ago
6

Arginine is one of the amino acids; it is used in the biosynthesis of proteins. Analysis revealed that a sample of arginine was

41.368 % carbon, 8.101% hydrogen, 32.162 % nitrogen and 18.369% oxygen. What is the empirical formula of arginine?
Chemistry
1 answer:
Anastasy [175]1 year ago
4 0

The empirical formula of arginine is C3H7N2O2.

<h3>What is the empirical formula of arginine?</h3>

Now we know that the empirical formula is the formula that shows the ratio of the atoms present in the compound.

Now we know that;

C - 41.368/12    H -  8.101/1     N -  32.162/14    O - 18.369/16

C - 3.445           H -  8.101      N - 2.297           O - 1.148

Dividing through by the lowest ratio;

C - 3.445/1.148    H -  8.101/1.148    N - 2.297/1.148   O - 1.148/1.148

C - 3                   H - 7                     N - 2                   O - 2

C3H7N2O2

Learn more about arginine:brainly.com/question/13589467

#SPJ1

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How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?
Anna [14]

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1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

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8 0
2 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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