The correct electron configuration for magnesium is: 1s 22s 22p 63s 3 True False

Answers

Answer 1

Answer:

Answer:

False

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or $1s^22s^22p^63s^2$

There are 2 valence electrons of magnesium.

Only the valence electrons are shown by dots in the Lewis structure.

As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.

Given that the electronic configuration is:- $1s^22s^22p^63s^3$ .

Orbital s cannot accommodate 3 electrons and also in magnesium it has $3s^2$ . Hence, the statement is false.

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I just took the test it is A hope this helps :)

What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.

Answers

Answer: The mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Explanation:

We are given:

Molality of calcium chloride = 3.35 m

This means that 3.35 moles of calcium chloride are present in 1 kg or 1000 g of water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18.02 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=(1000g)/(18.02g/mol)=55.49mol$

Total moles of solution = [3.35 + 55.49] = 58.84 moles

Mole fraction of a substance is given by:

$\chi_A=(n_A)/(n_A+n_B)$

For calcium chloride:

$\chi_(CaCl_2)=(n_(CaCl_2))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(CaCl_2)=(3.35)/(58.84)=0.057$

For water:

$\chi_(H_2O)=(n_(H_2O))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(H_2O)=(55.49)/(58.84)=0.943$

Hence, the mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Answer:

49.3% water

Explanation:

A 51.9g sample of iron, which has a specific heat capacity of 0.449·J·g?1°C?1, is put into a calorimeter (see sketch at right) that contains 300.0g of water. The temperature of the water starts off at 19.0°C. When the temperature of the water stops changing it's 20.3°C. The pressure remains constant at 1atm. Calculate the initial temperature of the iron sample. Be sure your answer is rounded to 2 significant digits.

Answers

Answer:

the initial temperature of the iron sample is Ti = 90,36 °C

Explanation:

Assuming the calorimeter has no heat loss to the surroundings:

Q w + Q iron = 0

Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )

Assuming Q= m*c*( T- Tir)

mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

Note :

- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C

- We assume no reaction between iron and water

Final answer:

To calculate the initial temperature of the iron sample, use the equation q = m * c * T, where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature which is 90.36 °C

Explanation:

To calculate the initial temperature of the iron sample, we can use the equation:

q = m * c * T

Where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, andT is the change in temperature. In this case, we know the mass of the iron sample, the specific heat capacity of iron, and the change in temperature of the water. By rearranging the equation, we can solve for the initial temperature of the iron sample.

Thus,

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

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The quantity of mass of an object contained within its volume is a measure of

Answers

the answer is density

the quantity of mass of an object contained within its volume is a measure of density

What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.

Answers

Answer:

K₂O

Explanation:

Given parameters:

Mass of K = 36.7g

Mass of O = 7.51g

Unknown:

Empirical formula of the compound

Solution:

The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.

To find the empirical formula, follow this process;

Elements K O

Mass 36.7 7.51

Molar

mass 39 16

Number of

moles 36.7/39 7.51/16

0.94 0.47

Divide by

the smallest 0.94/0.47 0.47/0.47

2 1

Empirical formula is K₂O

Final answer:

The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.

Explanation:

To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.

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Indicate the number of protons, neutrons, and electrons in each of the following species: 15N7, 33S16, 63Cu29, 84Sr38, 130Ba56, 186W74, 202Hg80

Answers

The number of protons neutrons, and electrons in each of the following species given are below;

What is the atomic number?

The total number of protons present in an atom is known as the atomic number of that atom. The atomic number has no correlation either with the number of neutrons or the number of electrons present inside an atom.

15N7 ⇒ 7 electrons, 8 neutrons, 7 protons

33S16 ⇒ 16 protons, 16 electrons, 17 neutrons

63Cu29 ⇒ 29 electrons, 34 neutrons,29 protons

84Sr38 ⇒ 38 electrons, 46 neutrons,38 protons

130Ba56 ⇒ 56 electrons, 74 neutrons,56 protons

186W74⇒ 74 electrons, 112 neutrons,74 protons

202Hg80 ⇒ 80 electrons, 122 neutrons ,80 protons

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Answer:

1. 7 protons, 7 electrons, 8 neutrons