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poizon [28]
4 years ago
13

2. What is the electron configuration of an electrically neutral atom of magnesium?

Chemistry
1 answer:
alekssr [168]4 years ago
7 0
Are u talking about electron sublevel config or where the electrons show in the "rings" of the atom
You might be interested in
However, when the temperature increases, the solubility of gaseous solutes will _______?
babunello [35]

Answer:

Decrease

Explanation:

Since the speed in which the gas molecules are faster as they are heated, they fly around in the container and logically, it is harder to insert a moving object into water than something more stationary or slower.

7 0
3 years ago
What is the mass, in grams, of 1.4 mol W
algol [13]
Atomic mass W = 183.84 u.m.a

1 mole --------- 183.84
1.4 moles ---- ?

1.4 x 183.84 / 1 = 257.376 g

hope this helps!
3 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
The pressure inside of a sealed tube if you raise the temperature go ______..
Helen [10]

Answer:

The pressure will increase due ot expnasion of gasses in a closed sealed tube tube .

Explanation:

4 0
3 years ago
Read 2 more answers
The colligative molality of an unknown aqueous solution is 1.56 m.
yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- <u>Vapor pressure lowering</u>

ΔP = P° . Xm . i

- <u>Boiling point elevation</u>

ΔT = Kb . m . i

-<u> Freezing point depressión</u>

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are  specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0
4 years ago
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