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1 year ago

Which substance can not be broken down by a

Chemistry

1 answer:

Tomtit [17]1 year ago

7 0

Answer: (3) silicon

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Silicon (Si)

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: ethane $(C_2H_6)$ , propanone $(CH_3COCH_3)$ and water $(H_2O)$

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All amino acids that are found in proteins, except for proline, contain a(n); Group of answer choices amino group carbonyl group

Jobisdone [24]

Answer:

amino group

Explanation:

There are twenty (20) amino acids in nature. Generally, each amino acid is structurally made up of a central carbon atom called alpha carbon attached to a hydrogen, carboxylic acid group (-COOH) and an amine group (-NH2). However, one particular amino acid called PROLINE posseses an exception to this.

Proline, which is the only cyclic amino acid, is also the only amino acid that forms a secondary amine group i.e. loss of hydrogen atoms in its amine group when in a protein structure. This means that when in a protein, PROLINE does not have an AMINE GROUP.

7 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

2 years ago

While hiking in the woods, you see a brown, rotting apple lying on the ground. What changes is the physical properties of the ap

mojhsa [17]

Answer:

The change in color.

Explanation:

The apple turn brown in color because of the oxidation process. When the oxygen and water molecules in air react with it, oxidation take place. The oxidation process is very efficient in ambient temperature.

For example, if the peal off apple is placed into the refrigerator it take a time to got oxidize and turn brown, but if it is placed in room temperature it quickly turn brown.

when oxygen is react with peel off apple , it trigger the polyphenol oxidase enzyme to oxidize the phenolic compound and quinones are formed which then react with amino acids and produced brown color.

3 0

1 year ago

1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?

larisa [96]

Answer:

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

7 0

1 year ago

Read 2 more answers

A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f

swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0

2 years ago