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Zepler [3.9K]
4 years ago
15

A protein has the amino acid sequence

Chemistry
1 answer:
kherson [118]4 years ago
6 0

Answer:

Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments

Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments

Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)

Explanation:

Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL

Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL.  All produced in three cut sites.

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu
tatyana61 [14]

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

5 0
3 years ago
The diameter of carbon atom is 0.000000000 154m. What is the number expressed in scientific notation
kogti [31]
1.54×10 −10

one and fifty four-hundreths times ten to the power of negitiive 10
7 0
3 years ago
Read 2 more answers
Which group would Ns2np2 go into
FrozenT [24]

Answer:

The group 15 elements: the pnicogens

Explanation:

The group 15 elements, nitrogen, phosphorus, arsenic, antimony and bismuth, all have the general valence shell electronic configuration ns2np3. They can all exist in the +3 or +5 oxidation state, with the +3 state increasing in stability as we move vertically down the group.

3 0
3 years ago
Read 2 more answers
A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe
nikitadnepr [17]

Answer:

\large \boxed{34.2\, ^{\circ}\text{C}}

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:  

For the metal:

m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

For the water:

m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}

\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}

\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

3 0
3 years ago
What property of water enables ice wedging to break rock?
seropon [69]

Answer:

c

Explanation:

hope I helped

hope im corrext

6 0
3 years ago
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