Answer: The child isotope has an atomic mass of 206.
Explanation:
Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.
General representation of an element is given as:
where,
Z represents Atomic number
A represents Mass number
X represents the symbol of an element
General representation of alpha decay :
Answer:
Sn^2+(aq)/Sn^3+(aq)//F2(g)/2F-(aq)
Explanation:
In writing the shorthand notation for an electrochemical cell, the oxidation half cell is shown on the left hand side and the reduction half cell is shown on the right hand side. The oxidation half equation reflects electron loss while the reduction half equation reflects electron gain.
Answer:
D. 3.3 x 10^24
Explanation:
In a number representing a measurement, the significant figure are those digits that have a meaningful contribution to specify the resolution of the measurement.
When a number is written in scientific notation, it is written as:
where
m is the base
n is the exponent
When a number is written in this form, the number of significant figures is the number of digits contained in .
In this problem, our number is initially written as
We see that here "m" contains 4 digits, so here we have 4 significant figures.
In order to reduce it to 2 significant figures, we have to rewrite 3.311 as 3.3, therefore the new number will be:
Which is option D.
Q = mCΔT
Q is heat in Joules, m is mass, C is the specific heat of water, delta T is the change in temperature
Q = (35g)(4.18)(35 degrees) = 5121 Joules or 5.12 kJ required
Answer:
0.33%
Explanation:
From the question given above, the following data were obtained:
Experimental value = 9.842 m/s²
Accepted value = 9.81 m/s²
Percentage error =?
The percentage error in the calculation can be obtained as follow:
Percentage error = |Experimental – Accepted| / Accepted value × 100
Percentage error = |9.842 – 9.81| / 9.81 × 100
Percentage error = 0.032 / 9.81 × 100
Percentage error = 0.33%
Therefore, the error in the calculation is 0.33%