If we place the like poles of two magnets facing each other what will happen?

It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons co

kotegsom [21]

<span>It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons configurations for the rest of the p-block elements in the periodic table beacause from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy level. </span>

3 0

2 years ago

Read 2 more answers

In addition to DNA, what else is

Pepsi [2]

Answer: ribosomes and membranes.

Explanation:

4 0

2 years ago

Oxidation number of Mg3N2

Fudgin [204]

Answer:

n=-3

Explanation:

Mg = +2

3(+2) + 2N = 0

N = -3

8 0

3 years ago

How do I calculate the boiling point of 25g of calcium chloride in 120g of water

Veronika [31]

I think, either 100 degrees celsius (as of water)
56 degrees celsius (acid)
or 118.1 degrees celsius (acetone)

5 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

2 years ago