Silicon is the element having a mass of 28.09 g
<u>Explanation</u>:
- Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023
10^23 atoms. One mole of each element can produce one mole of compound.
- The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 10^23).
- Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023 10^23 atoms of an element are equal to the average atomic mass of that element.
Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to create oxygen and energy in the form of sugar.
The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
Answer:they are used to generate sound waves in the ultrasonic range
Explanation:transducers are used to generate sound waves in the ultrasonic range ,by turning electrical energy into sound
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
