Answer : The types of radiation known to be emitted by radioactive elements are, alpha particles, beta particles, or gamma rays.
Explanation :
Radioactive decay : It the process in which an unstable atomic nucleus loses energy by emitting the radiations like, alpha particles, beta particles, or gamma rays.
The naturally occurring radioactive elements are, radium, thorium, and uranium.
Alpha particle : It is also known as alpha radiation or alpha ray that consists of 2 protons and 2 neutrons that are bound together into a particle that is identical to the helium nucleus. It is produced in the process of alpha decay.
Beta particle : It is also known as beta radiation or beta ray. During the beta decay process, a high energy and speed electron or positron are emitted by the radioactive decay of atomic nucleus.
Gamma particle : It is also a gamma radiation or gamma ray that is arising from the radioactive decay of atomic nuclei. It has shortest wavelength waves and imparts high photon energy can pass through most forms of matters because they have no mass.
Answer:
The molality of the KCl solution is 11.8 molal
Explanation:
Step 1: Data given
Mol fracrion KCl = 0.175
Molar mass KCl = 74.55 g/mol
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mol fraction H2O
mol fraction H2O = 1 - 0.175 = 0.825
Step 3: Calulate mass of H2O
Suppose the total moles = 1.0 mol
Mass H2O = moles H2O * molar mass
Mass H2O = 0.825 * 18.02 g/mol
Mass H2O = 14.87 grams = 0.01487 kg
Step 4: Calculate molality
Molality KCl = 0.175 / 0.01487 kg
Molality KCl = 11.8 molal
The molality of the KCl solution is 11.8 molal
solution:
the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃
pH of the buffer solution is 4.7
the value of pKa of sodium bicarbonate is 6.37
the value of pKa of acetic acid is 4.7
calculate concentration of acetic acid by using the following forumula
pH=pKa+lag[salt]/[acid]
substitute the pH and Pka values in the formula.
4.7=4.7+log[salt]/[acid]
log[salt]/[acid]=0
thus, the concentration ratio of the salt and acid should be equal to each other.
Thus, concentration of sodium acetate is 0.05M
Concentration of sodium acetate= concentration of acid
= 0.05M
Volume of the buffer solution is 100mL
The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.
The chemical equation for neutralization of the weak base with strong can be represented as show as
CH₃COONa+HCL-->CH₃COOH+NaCL
Answer:
0.8078 Kg
Explanation:
Pressure of water = 0.15 MPa = 1.5 bar
At critical point of water ,temperature = 647 K=374°C
From the ideal gas equation
P×V= m×R×T
Let us assume volume = 1 m^3
1.5 x 105 x 1 = m x 287 x 647
m= 0.8078 kg
the fraction of mass of liquid at 25°C.
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
