The molar mass of X would be 63.55 g/mol and the chemical symbol of X would be Cu (copper).
From the equation of the reaction: F2+2XBr⟶Br2+2XF
The mole ratio of XBr to Br2 is 2:1
Mole of Br2 = mass/molar mass
= 0.7241/159.808
= 0.004531 mole
Mole of XBr = 2 x mole of Br2
= 2 x 0.004531
= 0.009062 mole
Molar mass of XBr = mass/mole
= 1.300/0.009062
= 143.456 g/mol
Since the molar mass of Br is 79.904, the molar of X would be:
143.456 - 79.904
= 63.55 g/mol
Hence, X would be Cu.
More on molar mass calculation can be found here: brainly.com/question/20691135
Answer:
39 mol AgNO3
Explanation:
We have the equation 4HNO3 + 3Ag -----> 3AgNO3 + NO + 2H2O
We want to calculate the number of silver nitrate (AgNO3) moles that would be produced from 52 moles of nitric acid ( HNO3 )
We can calculate this by using mole ratio as well as dimensional analysis.
The mole ratio of Silver nitrate to nitric acid based on the balanced equation is 3AgNO3:4HNO3.
Using this we can create a table: The table is attached.
Breakdown of the table.
The moles of nitric acid cancel out and we multiply 52 by 3/4 to get 39 moles of Silver nitrate.
Answer:
Solution Density of aluminium = 2.7 g/Cm 3 In kg/ m 3 = 27 × 1000 10 =2700 kg/ m 3
Explanation:
Not much of one
Answer:
Option B is correct. A nuclear alpha decay
Explanation:
Step 1
This equation is a nuclear reaction. So it can be an alpha decay or a beta decay
An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.
During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.
Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation that is showed.
A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.
Option B is correct. A nuclear alpha decay
The relation between the elements electro negativity, atomic radius, and ionization energy levels.
