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UNO [17]
3 years ago
5

Give the (Hawarth projection) cyclic

Chemistry
1 answer:
denpristay [2]3 years ago
8 0

Answer:

a) The cyclic structure of D-fructose is drawn in the attachment

b)The carbons in the ring are 4

c) The diagram is shown in the attachment

d)This can be classified as a reducing agent. This is because it contains aldehyde that are oxidized to carboxylic acid.

e) The D-fructose sugar has anomeric carbon which is an aldehyde or hemiacetal. In fructose, both anomeric carbons are in acetal functional groups

Explanation:

The explanations is attached.

Download pdf
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ziro4ka [17]

Answer:

Torrey's neighbour is incorrect because increase in kinetic energy is proportional to velocity.  If the velocity increases so will the object's kinetic energy.  Because the mass is constant, if the velocity increases, so does the kinetic energy.

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What does food provide to organisms?<br><br> A-homeostasis<br> B-shelter<br> C-energy<br> D-air
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Energy would be the correct answer
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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
Grace [21]
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
     V2 = 34.1 mL
3 0
3 years ago
Read 2 more answers
Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n
mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide (CO_2), nitrogen dioxide (NO_2), and no other substances. A small sample gives 2.389 g CaO, 1.876 g CO_2, and 3.921 g NO_2 Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is CaCN_2

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of CO_2=1.876g

Mass of NO_2=3.921g

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 1.876 g of carbon dioxide, \frac{12}{44}\times 1.876=0.5116g of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, \frac{14}{46}\times 3.921=1.193g of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, \frac{40}{56}\times 2.389=1.706g of calcium will be contained.

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium =\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles

Moles of Nitrogen = \frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles

  • <u>Step 2:</u> 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 0.0426 moles.

For Calcium = \frac{0.0426}{0.0426}=1

For Carbon = \frac{0.0426}{0.0426}=1

For Nitrogen = \frac{0.0852}{0.0426}=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is CaCN_2

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3 years ago
Based on the chart, which species are most closely related?
pogonyaev

Answer:

Species B and C

Explanation:

7 0
2 years ago
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