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marishachu [46]
2 years ago
14

What assumption is being made if scientists conclude that aspartic acid was formed by the prebiological synthesis in the passage

?
A.Aspartic acid is unstable at temperatures below 150°C.
B.All of the malic acid underwent the dehydration reaction to form fumaric/maleic acid.
C.Compound A and cyanide were available on primitive Earth.
D.The reaction between ammonia and fumaric acid was catalyzed by the presence of water.
Chemistry
2 answers:
miss Akunina [59]2 years ago
8 0

Answer:

A....

Explanation:

almond37 [142]2 years ago
6 0

Answer:

A. Aspartic acid is unstable at temperatures below 150°C.

Explanation:

Aspartic acid  is an amino acid that is used in the production of proteins within  living cells.  It contains an amino group and a carboxylic acid, and can be synthesized by the body. It is a non-essential amino acid. The prebiological formation of aspartic acid occured at a certain temperature range. It was deduced from the passage that aspartic acid is unstable at temperatures below 150°C. The transamination of oxaloacetate in the human body gives rise to the formation of aspartate which is facilitated by the enzyme, aminotransferases. Aspartate also play a major role in the formation of urea by acting as a metabolite.

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How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO3 + CaC
nexus9112 [7]

Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with \frac{6.2}{2} = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

6 0
2 years ago
When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nac
Ann [662]

Solution:

After the reaction of mixture is worked-up Washing three times the organic  with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

5 0
2 years ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
2 years ago
Read 2 more answers
A student is given a sample of CuSO4(s) that contains a solid impurity that is soluble and colorless. The student wants to deter
DochEvi [55]

Answer:

The impurity which is present in the solution of copper sulphate (CuSO4) is determined by the an instrument known as spectrophotometer.

Explanation:

Spectrophotometer is a device or an instrument which is used to determine the concentration of a chemical by measuring the detection of light intensity that is coming from the solution. If the solution of copper sulphate is checked through spectrophotometer, we can can determined or measure the amount of copper sulphate and the impurity in the solution.

3 0
2 years ago
Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh
Naya [18.7K]

Answer:

1.82 g   is the maximum mass of CuS.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

<u>For CuCl_2 : </u>

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of CuCl_2 :

Moles=0.500 \times {38.0\times 10^{-3}}\ moles

<u>Moles of CuCl_2  = 0.019 moles </u>

<u>For (NH_4)_2S : </u>

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of (NH_4)_2S :

Moles=0.600 \times {42.0\times 10^{-3}}\ moles

<u>Moles of (NH_4)_2S  = 0.0252 moles </u>

According to the given reaction:

CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}

1 mole of CuCl_2 reacts with 1 mole of (NH_4)_2S

So,  

0.019 mole of CuCl_2 reacts with 0.019 mole of (NH_4)_2S

Moles of (NH_4)_2S = 0.019 mole

Available moles of (NH_4)_2S = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, CuCl_2 is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of CuCl_2 gives 1 mole of CuS

0.019 mole of CuCl_2 gives 0.019 mole of CuS

Moles of CuS formed = 0.019 moles

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g   is the maximum mass of CuS.</u>

5 0
2 years ago
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