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Snezhnost [94]
3 years ago
9

How do the amino acids lysine and alanine combine to form a dipeptide?

Chemistry
1 answer:
rewona [7]3 years ago
7 0
A dipeptide is formed by the process of dehydration synthesis, which in this case allows two amino acids to come together and form a larger molecule plus water.
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What is the percent, by mass, of water in MgSO4.2H20
iris [78.8K]

Answer:51.1%

Explanation:

Mass percent : It is defined as the mass of the given component present in the total mass of the compound. Formula used : First we have to calculate the mass of  and . Mass of  = 18 g/mole Mass of  = 7 × 18 g/mole = 126 g/mole Mass of  = 246.47 g/mole Now put all the given values in the above formula, we get the mass percent of  in . Therefore, the mass percent of  in  is, 51.1%

8 0
2 years ago
For metalloids on the periodic table, how do the group number and the period number relate?
Studentka2010 [4]

Answer : The correct option is A.

Explanation :

Metalloid : Metaloids are the elements whose properties lie between the metals and non-metals.

There are six commonly elements which are boron, silicon, germanium, arsenic, antimony and tellurium. The rare elements are polonium and astatine.

The relation between the group number and period number of metalloid is that the lower the group number, the lower the period numbers and the metalloids are found in a diagonal moving down from left to right in the periodic table.

5 0
3 years ago
Read 2 more answers
What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?
bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0
3 years ago
Read 2 more answers
Help Please!!!!!!!!!!
Shkiper50 [21]

Answer:

L/EGFOU;T4444444444444444444444czgfryewi;adkb,SJJ>RL:IAO:YHSBRAGldOUSDHRIUITUER

Explanation:

DHFUIEY7RY8EFUIDJKJEUSDYRIFU8ERJFHJSX

4 0
3 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

3 0
3 years ago
Read 2 more answers
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