In general chemistry, isotopes are a group of substances that belong to the same element. An element is characterized in the periodic table by their atomic number, which is the number of protons in an atom. Therefore, these substances have the same atomic numbers, but differ in mass numbers. Mass number is the sum of the number of protons and neutrons in the nucleus of an atom.
To determine the atomic weight of an element, you take the average weight of all the existent isotopes of that said element. The calculation would require to multiply the exact mass of the isotope to its abundance. Then, sum them all up.
Atomic weight = 98(0.18) + 112(0.82)
Atomic weight = 109.48 amu
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
Answer:
The correct option is C.
Explanation:
Carbohydrates are one of the macro molecules that are consumed by living organisms. The end product of carbohydrate is glucose. Glucose is a very important fuel that the body cells used to produce energy, which they use to carry out their daily activities. Glucose is also known as blood sugar and it is the only fuel that living cells can use for the production of ATP. Other food macro molecules such as lipids and proteins can also be converted to glucose if there is a need for that. Glucose is always stored in the body in form of glycogen.
The statement given in option C about glucose is wrong because glucose is a monosaccharide and not a disaccharide.
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
Answer:
2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO
