The number in isotope platinum-194 stands for the <em>total </em>amount of protons and neutrons. In other words, this is the mass. =)
Answer:
Moles of BCl₃ needed = 0.089 mol
Explanation:
Given data:
Moles of BCl₃ needed = ?
Mass of HCl produced = 10.0 g
Solution:
Chemical equation:
BCl₃ + 3H₂O → 3HCl + B(OH)₃
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 36.46 g/mol
Number of moles = 0.27 mol
Now we will compare the moles of HCl with BCl₃.
HCl : BCl₃
3 : 1
0.27 : 1/3×0.27 = 0.089 mol
<u>Answer:</u> The chemical equation is written below.
<u>Explanation:</u>
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
The chemical equation for the combustion of ethyl chloride follows:
We are given:
When 4 moles of ethyl chloride is burnt, 5145 kJ of heat is released.
For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.
For an exothermic reaction, heat is getting released during a chemical reaction and is written on the product side
So, the chemical equation follows:
Hence, the chemical equation is written above.
Rydberg Eqn is given as:
1/λ = R [1/n1^2 - 1/n2^2]
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present
q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present
sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present
