Write the symbolic notation of an isotope of an element having 8 protons, 8 electrons, and 11 neutrons. Click on the “Templates”
button template button and make use of the "stacked super/subscript" button button for entering stackes super/subscripts for entering the mass number and atomic number of the isotope. Express your answer as an isotope.
1 answer:
Attached to this answer is the format of Isotope Notation that you can use for future reference. <em>(Please open)</em>
There are 8 Protons. The Atomic Number is the same number of an element's proton.
If you can see in the format, the mass number is calculated by adding the atomic number/protons and neutrons.
Mass number = 8 + 11
Mass number = 19
The image of the final answer is attached as well.
There are 8 Protons. The Atomic Number is the same number of an element's proton.
If you can see in the format, the mass number is calculated by adding the atomic number/protons and neutrons.
Mass number = 8 + 11
Mass number = 19
The image of the final answer is attached as well.
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Answer:
b
Explanation:
The reaction that is not a displacement reaction from all the options is
In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.
<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>
Correct option = b.
Answer: 3.36 L of ammonia gas
Explanation:
The balanced chemical reaction is:
According to stoichiometry :
3 moles of produce = 2 moles of
Thus 0.75 moles of will producee=
of
But as percent yield is 30 %, amount of ammonia produced =
According to ideal gas equation:
P = pressure = 1 atm
V = Volume = ?
n = number of moles = 0.15
R = gas constant =
T =temperature =
Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.