Answer:
Mg²⁵ = 10.00%
Mg²⁶ = 45.04%
Mg²⁴ = 44.96%
Explanation:
Given data:
Atomic mass of Mg²⁶ = 25.983
Atomic mass of Mg²⁵ = 24.986
Atomic mass of Mg²⁴ = 23.985
Abundance of Mg²⁵ = 10.00%
Abundance of Mg²⁶ = ?
Abundance of Mg²⁴ = ?
Solution:
Average atomic weight of Mg = 25.983 + 24.986+ 23.985 / 3
Average atomic weight of Mg = 74.954/3
Average atomic weight of Mg = 24.985 amu
Abundance of
Mg²⁵ = 10.00
Mg²⁶ = x
Mg²⁴ = 100- 10 - x = 90 -x
Formula:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
24.985 = (0.1×24.986)+(90-x×23.985) + ( x ×25.983 ) /100
24.985 = 249.86 + 2158.65 - 23.985x + 25.983x / 100
24.985 = 2408.51 + 1.998 x / 100
2498.5 = 2408.51 + 1.998 x
1.998 x = 2498.5 - 2408.51
1.998 x = 89.99
x = 89.99 /1.998
x = 45.04
Now we put the value of x:
Mg²⁵ = 10.00
Mg²⁶ = x (45.04)
Mg²⁴ = 90 -x (90 - 45.04 = 44.96)
Answer:
Mole fraction of Nacl is 0.173
Explanation:
we know that
where,
P
sol - the vapor pressure of the solution
χ solvent - the mole fraction of the solvent
P
∘
solvent - the vapor pressure of the pure solvent
This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at
25
°
C You can use an online calculator to find that the vapor pressure of pure water at 25 C is equal to about 23.8 torr
.
=0.827
Also we know that
This means that the mole fraction of sodium chloride is
χ_{Nacl}= 1-Χ_{water}
= 1-0.827 =0.173
Answer:
Option 2= Glucose
Explanation:
Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.
Facilitated diffusion:
it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.
Primary active transport:
The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.
Secondary active transport:
It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.
To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,
PV = nRT
If we are to retain constant the variable n and V.
The percent yield can therefore be solved through the following calculation,
n = (10.5 L)/(22.4 L) x 100%
Simplifying,
n = 46.875%
Answer: 48.87%
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even no = 3/6 = 1/2
no. less than 5 = 4/6 = 2/3
