On number 2 if the atom has 3 protons,4 neutrons and 3 electrons what is the charge of the atoms nucleus?

"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"

arlik [135]

Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

5 0

3 years ago

What mass of solid lead would displace exactly 234.6 liters of water?

eimsori [14]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The mass of solid lead would displace exactly 234.6 liters of water should be <span>2,674,440</span>

6 0

2 years ago

Read 2 more answers

For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that w

Sindrei [870]

Answer:

75 mg

Explanation:

We can write the extraction formula as

x = m/[1 + (1/K)(Vaq/Vo)], where

x = mass extracted

m = total mass of solute

K = distribution coefficient

Vo = volume of organic layer

Vaq = volume of aqueous layer

Data:

m = 75 mg

K = 1.8

Vo = 0.90 mL

Vaq = 1.00 mL

Calculations:

For each extraction,

1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62

x = m/1.62 = 0.618m

So, 61.8 % of the solute is extracted in each step.

In other words, 38.2 % of the solute remains.

Let r = the amount remaining after n extractions. Then

r = m(0.382)^n.

If n = 7,

r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg

m = 75 - 0.088 = 75 mg

After seven extractions, 75 mg (99.999 %) of the solute will be extracted.

5 0

3 years ago

Calculate the molar mass of a gas

spin [16.1K]

Answer:

MOLAR MASS = 32 g/mol

Explanation:

Condition of standard temperature and pressure(STP) are as follow:

Temperature = 273 K

Pressure = 1 atm (or 100000 Pa)

Here atm is atmosphere and Pa is Pascal

STP conditions arte used for measuring gas density and volume using Ideal Gas Law.Here 1 mole of ideal gas occupies 22.4 L of volume.

According toi Ideal Gas Equation :

PV = nRT

where P = pressure, n= number of moles, V = volume ,R= Ideal Gas Constant and T= temperature

$n=\frac{PV}{RT}$

From question:

V=280 ml = 0.28 L

P = 1 atm

R=0.08205 L atm/K mol

T=273 K

Putting values in above formula :

$n=\frac{1\times .280}{0.08205\times 273}$

n = 0.0125 moles

Now $n=\frac{given\ mass}{Molar\ mass}$

$Molar\ mass=\frac{given\ mass}{n}$

given mass = 0.4 g (given)

$Molar\ mass=\frac{0.4}{0.0125}$

On solving we get:

Molar mass = 32 g/mol

4 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

3 years ago