Answer:
The number of copper atoms 12.405 ×10²³ atoms.
The number of silver atoms 13.13 ×10²³ atoms.
Beaker B have large number of atoms.
Explanation:
Given data:
In beaker A
Number of moles of copper = 2.06 mol
Number of atoms of copper = ?
In beaker B
Mass of silver = 222 g
Number of atoms of silver = ?
Solution:
For beaker A.
we will solve this problem by using Avogadro number.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.
While we have to find the copper atoms in 2.06 moles.
So,
63.546 g = 1 mole = 6.022×10²³ atoms
For 2.06 moles.
2.06 × 6.022×10²³ atoms
The number of copper atoms 12.405 ×10²³ atoms.
For beaker B:
107.87 g = 1 mole = 6.022×10²³ atoms
For 222 g
222 g / 101.87 g/mol = 2.18 moles
2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms
You need the Avogadro's number. I can't remember exactly the calculation.
Answer:
Explanation:
Given that:
The Half-life of = is less than that of
Although we are not given any value about the present weight of .
So, consider the present weight in the percentage of to be y%
Then, the time elapsed to get the present weight of =
Therefore;
here;
= Number of radioactive atoms relating to the weight of y of
Thus:
--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of to be =
Then:
---- (2)
here;
= Number of radioactive atoms of relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:
replacing the half-life of =
( since )
∴
The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is
Answer:
Explanation: When solutions of potassium iodide and lead nitrate are combined?
The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. Chemical Equation Balancer Pb(NO3)2 + KI = KNO3 + PbI2. Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble. The reaction is an example of a metathesis reaction, which involves the exchange of ions between the Pb(NO3)2 and KI. The Pb+2 ends up going after the I- resulting in the formation of PbI2, and the K+ ends up combining with the NO3- forming KNO3. NO3- All nitrates are soluble. ... (Many acid phosphates are soluble.)