When an atomic nucleus emits an alpha particle it decay into an atom with atomic number 2 less and mass number 4 less. Thus Thorium 230 decay as follows.
230 90Th -------> 226 88Th + 4 2 He
thorium is in the atomic number 90 thus it during alpha decay it reduces to atomic number 88 while its 230 mass number reduces to 226
E. coli and Salmonella spp.
Answer:
2KClO3 -------> 2KCl + 3O2
Explanation:
First, in balancing a chemical reaction such as the one given in the question, you should understand that for an equation to be balanced, the number of atoms and ions on both sides of the equation that is the right and left side must be equal. This follows the law of conservation of mass which tells us that matter can neither be created nor destroyed but can be changed into another form.
Next is to begin balancing the equation by identifying and writing down the substances given:
KCl03 ---------> KCl + O2
Next is to count he number of the individual atoms on each side and find out if they are the same on both sides and if not you must follow the next step.
Add a corresponding number and use it to multiply the atoms involved
KClO3 ---------> KCl + O2
Oxygen is 3 on the left side and two on the other side, so we multiply the left hand side by 2 and the right hand side by 3
2KClO3 -----> KCl + 3O2
The potassium and Chlorine are no longer balanced, so you multiply the right had=nd side of KCl by 2.
2KClO3 -----> 2KCl + 3O2
The reactionis herefore balanced as both sides have equal number of atoms and ions.
Answer:
The mass of tin is 164 grams
Explanation:
Step 1: Data given
Specific heat heat of tin = 0.222 J/g°C
The initial temeprature of tin = 80.0 °C
Mass of water = 100.0 grams
The specific heat of water = 4.184 J/g°C
Initial temperature = 30.0 °C
The final temperature = 34.0 °C
Step 2: Calculate the mass of tin
Heat lost = heat gained
Qlost = -Qgained
Qtin = -Qwater
Q = m*c*ΔT
m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)
⇒with m(tin) = the mass of tin = TO BE DETERMINED
⇒with c(tin) = the specific heat of tin = 0.222J/g°C
⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C
⇒with m(water) = the mass of water = 100.0 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C
m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C
m(tin) = 163.9 grams ≈ 164 grams
The mass of tin is 164 grams
Answer:
The answer to your question is V2 = 1.82 l
Explanation:
Data
Volume 1 = 77 l
Pressure 1 = 18 mmHg
Volume 2 = ?
Pressure 2 = 760 mmHg
Process
Use Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (18 x 77) / 760
-Simplification
V2 = 1386 / 760
-Result
V2 = 1.82 l
