Isoflavones, quercetin and anthocyanins are all what type of phytochemicals

What is the relative atomic mass of an element

Gennadij [26K]

Answer:

20.2 amu.

Explanation:

Let A represent isotope ²⁰X

Let B represent isotope ²²X

From the question given above, the following data were obtained:

For Isotope A (²⁰X):

Mass of A = 20

Abundance (A%) = 90%

For Isotope B (²²X):

Mass of B = 22

Abundance (A%) = 10%

Relative atomic mass (RAM) =?

The relative atomic mass (RAM) of the element can be obtained as follow:

RAM = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

RAM = [(20 × 90)/100] + [(22 × 10)/100]

RAM = 18 + 2.2

RAM = 20.2 amu

Thus, relative atomic mass (RAM) of the element is 20.2 amu

4 0

3 years ago

Explain how materials are classified on the basis of electrical conductivity. Give two uses of these materials in day to day lif

vitfil [10]

Answer: On the basis of electrical conductivity, materials are classified as conductors, semi-conductors and non-conductors. Conductors : Substances through which electricity can easily pass through are known as conductors. ... For example, glass, wood are non-conductors. Wood is used to make tables, desks etc.

3 0

2 years ago

PLEASEEEE HELPPP I BEGGGG FOR HELPPP PLEASEE

belka [17]

Answer:

d

Explanation:

3 0

3 years ago

Read 2 more answers

Thunder and lightning make atoms true or false

guajiro [1.7K]

False, atoms can not be created that way

7 0

3 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0

3 years ago