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masha68 [24]
3 years ago
8

group of atoms that gains or loses electrons is called _______. A. a polar ion B. a non-polar ion C. a polyatomic ion D. a commo

n ion E. None of the Above
Chemistry
1 answer:
raketka [301]3 years ago
7 0

Answer: Option (C) is the correct answer.

Explanation:

An ion is defined as a specie which is formed when a neutral atoms tends to gain or lose an electron.

When a neutral atom gain an electron then it forms a negative ion whereas when a neutral atom tends to lose an electron then it forms a positive ion.

For example, a neutral fluorine atom on gaining an electron will form F^{-} ion. And, a sodium atom on losing an electron forms Na^{+} ion.

When a group of atoms form ions then it tends to form polyatomic ions.

Thus, we can conclude that group of atoms that gains or loses electrons is called a polyatomic ion.

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Consider the equation: 2NO2(g) N2O4(g). Using ONLY the information given by the equation which of the following changes would in
Reika [66]
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
7 0
3 years ago
Read 2 more answers
Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

7 0
2 years ago
Read 2 more answers
What is SO2 shape name?
muminat

Answer:Molecular Formula SO2

Hybridization Type sp2

Bond Angle 119o

Geometry V-Shaped or Bent

Explanation:

hope this helped <3

5 0
3 years ago
212 pb 82 is the isotope notation for iron. what is the atomic number, mass number, and number of both protons and neutrons?
o-na [289]

Hello!

The mass number in isotope notation is denoted A, the atomic number is denoted as Z, and the element is denoted as X.

In the given isotope, the mass of the isotope is 212 amu, and the atomic number is 82.

We know that the number of electrons, and protons are equal to the atomic number. Therefore, there are 82 protons. Also, to find the number of neutrons, we subtract the atomic number from the atomic mass.

212 - 82 = 130 neutrons

<u>Final answers</u>:  

  1. Atomic Number: 82
  2. Mass number: 212
  3. Number of Protons: 82
  4. Number of Neutrons: 130
5 0
4 years ago
The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially
Crank

Answer: The rate constant is 0.334s^{-1}

Explanation ;

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 4.26 min

a =  initial amount of the reactant  = 2.56 mg

a - x = amount left after decay process  = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}

k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}

k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}

Thus rate constant is [tex]0.334s^{-1}

4 0
3 years ago
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