All categories

9 months ago

What should a simplified model of a large molecule like glucose show?

1 answer:

5 0

Glucose is the simplest sugar and carbohydrate that provides energy. The simplified model of glucose (C₆H₁₂O₆) shows carbon, hydrogen, and oxygen atoms linked together.

<h3>What is glucose?</h3>

Glucose is an example of a carbohydrate macromolecule that is further classified as a monosaccharide. They are crystalline and fundamental units of carbohydrates.

The molecular formula of glucose is C₆H₁₂O₆ and the mass is 180.156 g/mol. It is an aldohexose that contains an aldehydic functional group. In its structure, there are six oxygen atoms, six carbon atoms, and twelve hydrogen atoms.

Therefore, the glucose molecule is composed of C, H, and O.

Learn more about glucose here:

brainly.com/question/2396657

#SPJ1

You might be interested in

68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi

Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we convert the given masses of reactants to moles, using their respective molar masses:

68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH

78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and Mg(NO₃)₂ is the limiting reagent.

Now we calculate how many Mg(OH)₂ are produced, using the moles of the limiting reagent:

0.528 mol Mg(NO₃)₂ * $\frac{1molMg(OH)_2}{1molMg(NO_3)_2}$ = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g

7 0

2 years ago

What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop

Gekata [30.6K]

This is a straightforward question related to the surface energy of the droplet.

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. 

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. 

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ 

The five smaller droplets need to have the same volume as the original. Therefore 

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. 

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. 

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². 

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ 
From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. 

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.

7 0

2 years ago

Determine the freezing point of a 0.51 molal

Elan Coil [88]

Answer:

-65.897°C.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation: ΔTf = Kf.m,

where, ΔTf is the depression in freezing point of chloroform solution.

Kf is the molal depression constant of chloroform (Kf = 4.70°C.kg/mol).

m is the molality of the solution (m = 0.51 m).

∴ ΔTf = Kf.m = (4.70°C.kg/mol)(0.51 m) = 2.397°C.

∴ The freezing point of the solution = (freezing point of chloroform) - ΔTf = (-63.5°C) - (2.397°C) = -65.897°C.

7 0

3 years ago

The unheated Gas in the above system has a volume of 20.0 L at a temperature of 25.0 C and a pressure of 1.00 atm. The gas is he

kipiarov [429]

Answer:

1.25 atm.

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 20L

Initial temperature (T1) = 25°C

Initial pressure = 1 atm

Final temperature (T2) = 100°C

Final volume (V2) = constant i.e remain the same

Final pressure (P2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 25°C = 25°C + 273 = 298K

Final temperature (T2) = 100°C = 100°C + 273 = 373K

Step 3:

Determination of the final pressure of the gas. This is illustrated below:

Since the volume is constant, the following equation, P1/T1 = P2/T2 will be used to obtain the final pressure of gas as follow:

P1/T1 = P2/T2

Initial temperature (T1) = 298k

Initial pressure = 1 atm

Final temperature (T2) = 373K

Final pressure (P2) =?

P1/T1 = P2/T2

1/298 = P2 /373

Cross multiply to express in linear form

298 x P2 = 1 x 373

Divide both side by 298

P2 = 373/298

P2 = 1.25 atm.

Therefore, the pressure of the heated gas is 1.25 atm.

3 0

2 years ago

An element has an atomic weight of 137 g/mol and a BCC crystal structure. If the volume density of the element is 2.2 g/cm3 what

kap26 [50]

Explanation:

Below is an attachment containing the solution to the question.

7 0

2 years ago