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Romashka [77]
3 years ago
9

Which of the following best identifies where long range order would be found

Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Answer:

There is no choices to choose from

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Write the symbol for every chemical element that has an atomic number less than 13 and atomic mass greater than 19.3 u
emmainna [20.7K]

Answer:

The symbol of each element is, Ne, Na, Mg, and Al.

Explanation:

Below is the list of elements that has an atomic mass of less than 19.3 u.

The atomic mass of Neon is 20.1797 u and the atomic number is 10.

The atomic mass of Sodium is 22.989769 u and the atomic number is 11.

The atomic mass of Magnesium is 24.305 u and the atomic number is 12.

The atomic mass of Aluminium is 26.981539 u and the atomic number is 13.

Here, the symbol of each element is, Ne, Na, Mg, and Al.

3 0
3 years ago
.Inherited traits are passed from parent to offspring by information coded in
Liono4ka [1.6K]

Answer:

Inherited traits are passed from parent to offspring by information coded in

the DNA or Deoxyribonucleic Acid

8 0
3 years ago
I need Helpppp with my chemistry while we are out of school. I don't have a good teacher and he's not helping since it's his las
9966 [12]

Answer:

im not sure but I hope this helps

Explanation:

I believe the equivalents is just the moles reactant/moles limiting reactant

water has a denisty of 1 g/mL. 1 L is 1000 ml so there are 1000g/L.

the molar mass of water is 18g/mol if you use the Liters in the equation above you can find the number of grams present. divide this number you found by 18 to find the moles.

now take the amount of the other reactant given and divide it by its own molar mass. this will give you the number of moles of that reactant.

divide the moles of water by the moles of the reactant and that is the equivalent.

to find the normality you take this number and divide it by the number of liters.

5 0
3 years ago
What does a catalyst do in the Haber process?
Ivenika [448]

Answer:

B

Explanation:

5 0
3 years ago
Read 2 more answers
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many
beks73 [17]

<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, A_0 = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}

where N is the no. of half life

Substituting the values,

\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}

\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0
3 years ago
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