Answer:
m = 20.9 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:
Regards!
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Answer:
0.0084
Explanation:
The mole fraction of BaCl₂ (X) is calculated as follows:
X = moles BaCl₂/total moles of solution
Given:
moles of BaCl₂ = 0.400 moles
mass of water = 850.0 g
We have to convert the mass of water to moles, by using the molecular weight of water (Mw):
Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol
moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol
The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):
total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol
Finally, we calculate the mole fraction:
X = 0.400 mol/47.6 mol = 0.0084
Three elements that are likely to have similar chemical and physical properties are
Answer:
Fluorine
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Periodic Trends
- Electronegativity - the tendency for an element to attract an electron to itself
- Z-effective and Coulomb's Law, Forces of Attraction
Explanation:
The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.
Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.
However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.
Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.
