Answer:
220.42098 amu
Explanation:
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
These are weighted averages.
So, we will take mass of one and multiply by abundance percentage that is provided and add them together.
In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
Answer:
Explanation:
The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.
Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.
Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.
Thus, OH⁻ and H⁺ ions will be neutralized by the reaction:
- OH⁻ (aq) + H⁺ (aq) ⇄ H₂O (l)
Which is strongly shifted to the right and has <em>neutral pH</em>.
Hence, you conclude that the approximate <em>pH of the solution is neutral.</em>
Answer:
The value of 
for this reaction at 1200 K is 4.066.
Explanation:
Partial pressure of water vapor at equilibrium = 
Partial pressure of hydrogen gas at equilibrium = 
Total pressure of the system at equilibrium P = 36.3 Torr
Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:
The expression of is given by:
The value of for this reaction at 1200 K is 4.066.
According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.
The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.
Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases
\frac{a}{b}=\frac{15}{35}=\frac{10}{y}
rearranging,
y=\frac{10\times 35}{15}=23.3
Thus, mass of b produced will be 23.3 g.
On the off chance that the red blood cells are smaller than ordinary, this is called microcytic anemia. The significant reasons for this sort are low-level iron, anemia, thalassemia.
