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3 years ago

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.

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Determine the soil texture for the following combinations of sand, silt, and clay.

JulijaS [17]

Answer:

a. Clay

b. Sandy Loam

c. Sandy Clay

d. Loamy Sand

Explanation:

Based on the soil texture triangle, the soil texture for the following combination of sand, silt and clay are;

a. Sand 20 percent; Silt 20 percent;Clay 60 percent: ____ CLAY

b. Sand 60percent; Silt 30percent; Clay 10percent: ___ SANDY LOAM

c. Sand 50percent; Silt 10percent;Clay 40 percent: ____ SANDY CLAY

d. Sand 80 percent; Silt 15 percent; Clay 5percent:_____ LOAMY SAND

7 0

3 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago

Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce

kkurt [141]

Explanation:

The given data is as follows.

P = 3 atm

= $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$

= $3.03975 \times 10^{5} Pa$

$V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$ (as 1 L = 0.001 $m^{3}$ ),

$V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

W = $P \times \Delta V$

or, W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

W = $P \times (V_{2} - V_{1})$

= $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

= 1823.85 Nm

or, = 1823.85 J

As internal energy of the gas $\Delta E$ is as follows.

$\Delta E$ = Q - W

= 800 J - 1823.85 J

= -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0

3 years ago

An electron ______ is an area around the nucleus of an atom where an electron is likely to be found.

jek_recluse [69]

Answer:

orbital

hope this helps! <3

4 0

2 years ago

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur

egoroff_w [7]

Answer: 6.26atm

Explanation:Please see attachment for explanation

7 0

3 years ago

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