All categories

1 month ago

How many molecules of glucose are in 1 l of a 100 mm glucose solution?

Chemistry

1 answer:

castortr0y [3K]1 month ago

6 0

The result is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of the glucose solution.

V(glucose) = 1 L; volume of the glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; quantity of substance.

N(glucose) = n(glucose) · Na (Avogadro's number).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

You might be interested in

How many molecules are in 0.25 grams of dinitrogen pentoxide?

eduard [2782]

To determine the answer, you need to understand the formula for converting grams to moles, which will then lead you to the number of molecules.
The result is 2 moles of N2O5. The process is as follows:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
Thus, the final answer is 2 molecules.

4 0

1 month ago

Read 2 more answers

Why is it important to have regular supervision of the weight and measurements in the market

lions [2927]

Answer:

Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.

Explanation:

4 0

2 months ago

"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2795]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

6 0

2 months ago

Apex is a new internet streaming service. Read their advertisement from the local newspaper. Apex is the newest, fastest streami

eduard [2782]

Answer:

B. Apex provides more channels per dollar compared to certain other streaming platforms.

Explanation:

Just completed the Quiz on Edgenuity.

7 0

1 month ago

Read 2 more answers

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

Tems11 [2777]

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

7 0

1 month ago