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2 years ago

Please answer

Chemistry

1 answer:

timofeeve [1]2 years ago

5 0

Answer:

1) petroleum, mixture, not found on periodic table

2) hydrogen and helium, mixture, contains "and"

3) iron sulphide, compound, ends with "ide"

4) carbon dioxide, compound, ends with "ide"

5) salt water, mixture, not found on periodic table

6) milk and sugar, mixture, contains "and"

7) silver nitrate, compound, ends with "ate"

8) sodium carbonate, compound, ends with "ate"

9) carbon electrode, mixture, not found on periodic table

10) astatine, element, found on periodic table

11) argon, element, found on periodic table

12) oxidane, compound, contains "ane"

13) iron and sulphur, mixture, contains "and"

14) methane, compound, contains "ane"

Explanation:

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Can you draw the peptide structure at pH 7.0 (including peptide bonds)

ELEN [110]

Answer:

See image below

Explanation:

The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.

Overall, ionizable groups in this peptide are:

Amino Group of the Tryptophan (pKa = 9.39)

SH group of the Cysteine (pKa = 8.18)

Secondary amine of the Histidine (pKa = 6.00)

Carboxylic Group of the Histidine (pKa = 1.82)

Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.

7 0

3 years ago

Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

ollegr [7]

Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0

3 years ago

What is the total number of orbitals found in the second energy level? openstudy?

Setler79 [48]

On the second shell there are two individual subshells:
The "s" subshell has only 1 orbital with max. two electrons spinning around; and the so-called "p" subshell has 3 orbitals with max. 6 electrons (2 on each!)
In total, there are four orbitals with 8 revolving electrons on the second shell.
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3 years ago

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Which reactant (X) is missing from the equation shown?<br> X+ PO4 → H3PO4

insens350 [35]

Answer:

Explanation:

b/c the polyatomic PO4¯³

have 3 valance number

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2 years ago

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kvv77 [185]

It should be the answer b but I’m not to sure

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3 years ago