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2 years ago

Part A

Chemistry

1 answer:

PilotLPTM [1.2K]2 years ago

6 0

Answer:

4200ml

Explanation:

Converting 3.1kg to g

3.1*1000= 3100g

Since density = mass/volume, then

volume = mass/ density

Therefore volume = 3100/0.74

= 4189.2ml

converting it to two significant figures

= 4200ml

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How does the numbers valence electrons in oxygen compare to the valence electrons in hydrogen?

elena-s [515]

Answer:

Determine the total number of valence electrons in the molecule or ion.

Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons

Explanation:

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3 years ago

This dipeptide is designated as: there is a structure for h3nchcnhchco with ch2oh group attached to the first (from left to righ

oksian1 [2.3K]

A dipeptide is formed when 2 amino acids are joined together by a peptide linkage.

Each amino acid molecule contains an amino group ( -NH₂) and a carboxyl ( -COOH) group. During peptide formation, a carboxyl group of one amino acid reacts with amino group of the other , resulting in the formation of a linkage known as "peptide linkage". A water molecule is also eliminated in this reaction.

A general peptide formation reaction can be written as

$NH_{2}-CH_{2}-COOH + H_{2}N-CH_{2}-COOH \rightarrow NH_{2}-CH_{2}-CONH-CH_{2}-COOH +H_{2}O$

In the above reaction, we can see a peptide linkage ( -CONH-) getting formed between 2 amino acid molecules.

In the given molecule the peptide linkage is shown by drawing a box around it. Please refer to the attached image.

If we want to know the amino acid molecules from which this peptide is formed, we will simply break the CO-NH bond.

We will add -OH group to CO part and -H to NH part to get 2 amino acid molecules as shown in the picture.

The two amino acids thus formed are Serine & Cysteine.

Serine is designated as "Ser" and Cysteine is designated as "Cys"

Therefore the given dipeptide is designated as "Ser-Cys"

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3 years ago

What is the percent composition of N H S O in (NH4)2SO4

Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

- 1 mole of S (32g/mol) = 32g

- 4 moles of O (16g/mol) = 64g

3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

$\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}$ $\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}$ $\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}$ $\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\ \\ \end{gathered}$

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

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1 year ago

Why do northern lights appear in different colors? Because charged particles of solar wind ignite different gases in Earth's atm

Fofino [41]

Answer:

It is A. Because charged particles of solar wind ignite different gases in Earth's atmosphere.

Explanation:

Since the solar wind from the sun is too radioactive for humans (they would die), once the charged particles hit the earth's atmosphere it shows its color. Every element has its own color and once it hits the atmosphere it really starts to show.

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3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$