What does the number next to isotopes signify?

Answers

Answer 1

Answer: The number next to isotopes signifies the sum of the number of protons plus the number of neutrons in the nucleus of an atom. An isotope is a version of an atom that has the same number of protons but different numbers of neutrons. Isotopes are very important in nuclear chemistry. A proton is a is a particle in the nucleus of an atom that has a positive charge. A neutron is also a particle in the nucleus of an atom, but it does not have an electric charge.

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What is the [H30+) in a solution with (OH'] = 2.0 x 10-3 m?2.0 x 10-3M5.0 x 10-11M2.0 x 10-12M5.0 x 10-12 M1.0 10-12 M

How much energy does a 9 x 10-8 m wavelength photon have?A. 5.96 x 10-41 j

B. 2.21 x 10-18 j

C. 7.36 x 10-27

D. 1.988 x 10-49 J

Answers

Answer:

Explanation:

From the given formukar for energy of a photon, we can obtain the energy of the photon as shown in the image. The meaning of the parameters in the equation are also shown in the image.

Answer: 2.21 x 10^-18 j

Explanation: a pex

Is biting an apple a physical or chemical change?

Answers

It would be physical because you are not using chemicals to make this change happen you r taking a bite out if the apple which is causing it to be a physical reaction

the answer is Physical because you are taking a bite out of the apple

Which of these did your answer include?Each hydrogen atom is bonded to the oxygen atom with a single bond.
The oxygen atom has two lone pairs.
The molecule is bent.
Oxygen is more electronegative than hydrogen.
Each O-H bond is polar.
The molecule is asymmetrical.
The asymmetry and the polar bonds produce an overall molecular dipole.
The oxygen atom has a partial negative charge, and the hydrogen atoms have a partial positive
charge.

Answers

The O2 atoms have a covalent bonding and are attracted to the electrons more strongly.

The sharing of the electrons gives them water molecules a slightly negative change near the O2 atom. A slight positive change near the H2.

Hence the option D is correct.

Learn more about the did your answer includes.

brainly.com/question/15913753.

The oxygen atom has a partial negative charge and the Hydrogen atoms have a partial positive charge

Which structures are found in both prokaryotic cells and eukaryotic cells?Choose all answers that are correct.

A.
cytoplasm

B.
flagella

C.
pili

D.
plasma membrane

E.
ribosomes

Answers

it is both D and E because both are found in most cells

What three parts make up a single nucleotide

Answers

A single nucleotide is made up of the three parts known as phosphoric acid, deoxyribose and a nitrogenous base

What three parts are in a single nucleotide ?

The part of the nucleotide that determines the genetic information is the Nitrogenous base. There are four different types of nitrogenous bases: adenine (A), guanine (G), cytosine (C), and thymine (T).

The part of the nucleotide that provides the structure and stability of the molecule. The sugar in DNA is deoxyribose, while the sugar in RNA is ribose.

The Phosphate group is part of the nucleotide that provides energy and helps to link nucleotides together. It is also known as phosphoric acid.

Find out more on nucleotides at brainly.com/question/14067588

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Hello!

A single nucleotide is composed of phosphoric acid, deoxyribose and a nitrogenous base (adenine, guanine, thymine or cytosine).

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your disposal, you have 5-mL and 10-mL transfer pipets and volumetric flasks of sizes 100, 250, 500, and 1000mL. Which of the following serial dilutions will give you the 200.0μM solution?

Answers

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

$500mM = 500mM * ((1M)/(1000M))= 0.5M$

Consider the following serial dilutions.

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

$M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(500 mL)= 5 * 10^(-3)M$

10 mL of this solution is diluted to 250 ml

$M_(final)= (M_(2)V_(2))/(V_(final))= ((5 * 10^(-3)M)(10.0mL))/(250 mL)= 2 * 10^(-4)M$

Convert μM :

$2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

$M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(100 mL)= 2.5 * 10^(-2)M$

10 mL of this solution is diluted to 1000 ml

$M_(final)= (M_(2)V_(2))/(V_(final))= ((2.5 * 10^(-2)M)(10.0mL))/(1000 mL)= 2.5 * 10^(-4)M$

Convert μM :

$2.5 * 10^(-4)M = (2.5 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 250 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

$M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(100 mL)= 0.05M$

5 mL of this solution is diluted to 1000 ml

$M_(final)= (M_(2)V_(2))/(V_(final))= ((0.05M)(5mL))/(1000 mL)= 0.25 * 10^(-4)M$

Convert μM :

$0.25 * 10^(-4)M = (0.25 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 25 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

$M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5mL))/(250 mL)= 0.01M$

10 mL of this solution is diluted to 500 ml

$M_(final)= (M_(2)V_(2))/(V_(final))= ((0.01M)(10mL))/(500 mL)= 2 * 10^(-4)M$

Convert μM :

$2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

$M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(250 mL)= 0.02M$

10 mL of this solution is diluted to 1000 ml

$M_(final)= (M_(2)V_(2))/(V_(final))= ((0.02M)(10mL))/(1000 mL)= 2 * 10^(-4)M$

Convert μM :

$2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

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