All categories

2 years ago

An element with 11 neutrons, 11 protons, and 11 electrons would have a mass number o?

Chemistry

1 answer:

vova2212 [387]2 years ago

4 0

The mass number of element is 22.

The mass number is the sum of number of electrons/proton and number of neutron .

Mass number = number of proton/electron + number of neutrons

If an element with 11 neutrons, 11 protons, and 11 electrons then mass number is calculated as,

Mass number = number of proton/electron + number of neutrons

Mass number = 11 + 11 = 22

learn about element

brainly.com/question/14347616

#SPJ4

You might be interested in

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

8 0

3 years ago

What do dipole-dipole forces do?

g100num [7]

An ion-dipole force is a type of intermolecular force in which forces of attraction or repulsion occur between neighboring ions, molecules or atoms.

3 0

3 years ago

Read 2 more answers

Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The

Maksim231197 [3]

Answer:

$\large\boxed{\large\boxed{24.6kJ}}$

Explanation:

1. Energy to heat the liquid water from 55ºC to 100ºC

$Q=m\times C\times \Delta T$

m = 10.1g

C = 4.18g/JºC

ΔT = 100ºC - 55ºC = 45ºC

$Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J$

2. Energy to change the liquid to steam at 100ºC

$L=\lambda \times n$

λ = 40.6kJ/mol

n = 10.1g / 18.015g/mol = 0.5606mol

$L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J$

3. Total energy

$1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ$

7 0

3 years ago

The pressure of a sample of helium in a 200. ml container is 2.0 atm. If the helium is compressed to a pressure of 40. atm witho

Crank

Answer:

$V_2=10mL$

Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

$P_2V_2=P_1V_1$

In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

$V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{2.0atm*200.mL}{40.atm}\\\\V_2=10mL$

Regards!

5 0

3 years ago

what is the concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 m dextrose solution to 25 ml using a 25 ml

allsm [11]

The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.

Concentration is defined as the number of moles of a solute present in the specific volume of a solution.

According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.

M₁V₁=M₂V₂

Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml

Rearrange the formula for M₂

M₂=(M₁V₁/V₂)

Plug all the values in the formula

M₂=(1.0M×14 ml/25 ml)

M₂=14 M/25

M₂=0.56 M

Therefore, the concentration of a dextrose solution after the dilution is 0.56M.

To know more about dilution

brainly.com/question/18566203

#SPJ4

6 0

1 year ago