Chemistry College

Which of the following contains a polyatomic ion?

sodium iodide

carbon monoxide

Iron II oxide

ammonium chloride

Answers

Answer 1

Answer:

i think its oxide not too sure

Answer 2

Answer:  

 Its should be ammonium chloride.

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Which Of The Following Contains A Polyatomic Ion?sodium Iodidecarbon MonoxideIron II Oxideammonium Chloride

Related Questions

Which rule states that cracks on glass tend to form at a certain angle on the opposite side of the polnt of Impact?
The
rule states that cracks tend to form at a (n)
angle on the opposite side from the point of Impact.

Answers

Answer:

The 3R rule states that cracks tend to form at a (n) right angle on the opposite side from the point of impact.

Explanation:

Which statement about Niels Bohr's atomic model is true?
O Higher orbits have lower energies.
O Each orbit has a specific energy level.
O Electrons can exist in any energy level.
O Orbits close to the nucleus have no energy.

Answers

Answer:

O Each orbit has a specific energy level.

Explanation:

Neils Bohr put forward his own model of the atom based on the quantum mechanics originally developed by Planck.

He assumed the Rutherford's model and suggested that the extranuclear part consists of electrons in specific spherical orbits around the nucleus.

The orbits/energy level are the permissible through which the electrons can move through.

His concept is based the concept that the electron can move round the nucleus in certain permissible orbitss

which would be a stronger acid H3PO4 or H3PO2​

Answers

Answer:

h3po2

Explanation:

Answer:

i think it is H3PO4

Explanation:

How many orbitals in an atom can have each of the following designations:
(a) 1s;
(b) 4d;
(c) 3p;
(d) n=3?

Answers

Answer:

(a) 1s; has one orbital

(b) 4d; has five orbitals

(c) 3p; has three orbitals

(d) n=3 has nine orbitals

Explanation:

Electrons in an atom are always in constant motion, making it hard to predict there exact position. However, the most probable locations electrons can be be found are described with the terms shells, subshells and orbitals. A shell contains subshells and orbitals are found within subshells. The shells are given names such as K, L, M, N, which correspond to the principal quantum numbers, n = 1, 2, 3, and 4 respectively. There are 4 major types of subshells that can be found in a shell. They are named as s, p, d, f. Each subshell is composed of several orbitals.

a. 1s; the s subshell has only one orbital. Therefore, the 1s subshell has one orbital

b. 4d; the d subshell has five orbitals. Therefore, the 4d subshell has five orbitals

c. 3p; the p subshell has three orbitals. Therefore, the 3d subshell has  three orbitals

d. n = 3; the shell with n = 3 has the following subshells, 3s, 3p, 3d.the number of orbitals will be 1 + 3 + 5 = 9 orbitals. Therefore, the number of orbitals in n = 3 is nine orbitals

A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C

Answers

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

For solution A

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

For solution B

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

For solution C

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

Learn more about serial dilution: https://brainly.com/question/2167827

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