Chemistry High School

Identify the element symbol
1s22s22p63s23p64s23d104p65s24d5

Answers

Answer 1

Answer:

Tc

Explanation:

You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!

Related Questions

copper has two stable isotopes,and , with masses of 62.939598 amu and 64.927793 amu, respectively. calculate the percent abundances of these isotopes of copper.

Answers

Considering the definition of atomic mass, isotopes and atomic mass of an element,  the isotope with masses of 62.939598 amu and 64.927793 amu have percent abundances of 69.80% and 30.20% respectively.

Definition of atomic mass

The atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

Definition of isotope

The same chemical element can have the same atomic numbers, but the number of neutrons is different. These atoms are called isotopes of the element.

Definition of atomic mass

The atomic mass of an element is the weighted average mass of its natural isotopes, taking into account the relative abundance of each of them.

Atomic mass of the element in this case

In this case, you know:

The first isotope has an atomic mass of 62.939598 amu and a percent natural abundance of x%. The second isotope has an atomic mass of 64.927793 amu and a percent natural abundance of (100 -x)%.Average mass of copper is 63.54 amu.

The percent abundance can be calculated as:

62.939598 amu× (x%÷100%) + 64.927793 amu× [(100-x)%÷100%]= 63.54 amu

Solving:

62.939598 amu× x+ 64.927793 amu× (1-x)= 63.54 amu

62.939598 amu× x+ 64.927793 amu× 1- 64.927793 amu× x= 63.54 amu

62.939598 amu× x- 64.927793 amu× x= 63.54 amu - 64.927793 amu× 1

-1.988195 amu× x= -1.387793 amu

x= (-1.387793 amu)÷ (-1.988195 amu)

x= 0.6980= 69.80%

Then, (100-x)%= (100 -69.80)%= 30.2%

Finally, the isotope with mass of 62.939598 amu has a percent abundance of 69.80% and the isotope with mass of 64.927793 amu has a percent abundance of 30.20%.

Learn more about average atomic mass:

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Put the following elements in order, with the element having the most valence electrons at the top of your list and the element with the fewest valence electrons at the bottom.swap_vertAstatine (At)swap_vertArsenic (As)swap_vertCalcium (Ca)swap_vertSodium (Na)swap_vertOxygen (O)swap_vertSilicon (Si)swap_vertAluminum (Al)

Answers

answer

1. Neon

2. Astatine

3. oxygen

4. Arsenic

5. Silicone

6. Aluminium

7. Calcium

8. Sodium

a reaction involving acetic acid has an enthalpy change of 55.7 kj/mol. what is the concentration of 27.90-ml of this acetic acid solution if the reaction it is involved in evolves 716.12-j of heat?

Answers

not gon find it here cuh

Explanation:

The concentration of the acetic acid solution is approximately 0.0128 mol/L, given the information provided.

To determine the concentration of the acetic acid solution, we need to use the given enthalpy change and the heat evolved in the reaction.

The enthalpy change (ΔH) is given as 55.7 kJ/mol, which represents the heat released or absorbed per mole of acetic acid involved in the reaction.

The heat evolved in the reaction is given as 716.12 J. However, we need to convert it to kilojoules to match the unit of the enthalpy change. There are 1000 J in 1 kJ, so 716.12 J is equal to 0.71612 kJ.

Now, we can set up a proportion to find the concentration of the acetic acid solution. The heat evolved is directly proportional to the moles of acetic acid involved in the reaction, which is in turn proportional to the concentration of the solution.

Let's assume the concentration of the acetic acid solution is c mol/L.

The proportion can be set up as:

ΔH (kJ/mol) / Heat evolved (kJ) = 1 mol / c L

Substituting the values we have:

55.7 kJ/mol / 0.71612 kJ = 1 mol / c L

Simplifying:

c = 1 mol / (55.7 kJ/0.71612 kJ)

c ≈ 0.0128 mol/L

Therefore, the concentration of the acetic acid solution is approximately 0.0128 mol/L, given the information provided.

Learn more about concentration from the link given below.

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