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skad [1K]
2 years ago
6

Glucose generally exists in ring (cyclic) form. a haworth projection shows the orientations of the hydroxyl groups and hydrogen

atoms on the ring. draw the α and β forms of glucose by moving the groups (h, oh, or ch2oh) to the appropriate positions. for the hydroxy groups, make sure the bond on the ring is bonded to the oxygen, not the hydrogen.

Chemistry
2 answers:
ladessa [460]2 years ago
8 0
Following are the required structures of glucose.

koban [17]2 years ago
8 0

The right answer is in the picture (alpha and beta glucose)

Glucose is the main source of energy in the body. It is a carbohydrate found in the diet, usually linked to more complex carbohydrates. It is once arrived in the intestine and degraded by enzymes during digestion that simple glucose is synthesized. The glucose is then converted into glycogen which is stored in the muscles and in the liver (reserve available for the entire body). In general and without any pathology, the blood glucose level is kept constant.

In the human body, both forms of glucose exist, with a predominance of beta-D-Glucose in the blood (66%).

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Calculate the ph of a solution that is 0.6 m in sodium acetate and 0.2 m in acetic acid. ka = 1.8 × 10−5 for acetic acid. 1. ph
Arte-miy333 [17]
      0.53 x 200ml = 106 ml of the pH 9.0 buffer + 94 ml of the pH 10 buffer gives the desired solution 
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4 0
1 year ago
Calculate the amount of heat necessary to raise the temperature of 135.0 g of water from 50.4°F to 85.0°F. The specific heat of
MAXImum [283]

Here we have to calculate the heat required to raise the temperature of water from 85.0 ⁰F to 50.4 ⁰F.

10.857 kJ heat will be needed to raise the temperature from 50.4 ⁰F to 85.0 ⁰F

The amount of heat required to raise the temperature can be obtained from the equation H = m×s×(t₂-t₁).

Where H = Heat, s  =specific gravity = 4.184 J/g.⁰C, m = mass = 135.0 g, t₁ (initial temperature) = 50.4 ⁰F or 10.222 ⁰C and t₂ (final temperature) = 85.0⁰F or 29.444 ⁰C.

On plugging the values we get:

H = 135.0 g × 4.184 J/g.⁰C×(29.444 - 10.222) ⁰C

Or, H = 10857.354 J or 10.857 kJ.

Thus 10857.354 J or 10.857 kJ heat will be needed to raise the temperature.

6 0
1 year ago
What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
german
When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

       = 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
7 0
2 years ago
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
1 year ago
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The answer is oxygen. (02)
4 0
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