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3 years ago

The diagram shows four different locations in an atom.

Chemistry

2 answers:

Alja [10]3 years ago

7 0

Answer:

1 and 4

Explanation:

In the figure attached, a plot of the problem is shown.

Location 1 and 4 are likely to have subatomic particles with the least mass because they are not in the nucleus, which is the location with most of the mass of the atom.

Ainat [17]3 years ago

5 0

Answer:

The answer is 1 and 4.

Explanation:

Mass is most concentrated in the nucelus of an atom. Therefore, if you are looking to find the area with the least mass, go outside of the nucelus. Points one and four are the furthest outside of the nucleus.

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How many liters of 3^He are present in 22 moles of gas?

Anna35 [415]

Look it up on goggle

8 0

3 years ago

Given the partial equation: NO3− Pb2 → NO2 Pb4 , balance the reaction in acidic solution using the half-reaction method and fill

Fed [463]

Answer : The balanced reaction in acidic solution is,

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

Explanation :

The given partial equation is,

$NO_3^-+Pb^{2+}\rightarrow NO_2+Pb^{4+}$

First we have to separate into half reaction. The two half reactions are:

$NO_3^-\rightarrow NO_2$

$Pb^{2+}\rightarrow Pb^{4+}$

Now we have to balance the half reactions in acidic medium, we get:

$NO_3^-+2H^++1e^-\rightarrow NO_2+H_2O$ ............(1)

$Pb^{2+}\rightarrow Pb^{4+}+2e^-$ ............(2)

Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2, we get

$2NO_3^-+4H^++2e^-\rightarrow 2NO_2+2H_2O$ ...........(3)

Now we have to add both the half reactions (2) and (3), we get the final balanced chemical reaction.

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

3 0

3 years ago

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

- Vapor pressure lowering

ΔP = P° . Xm . i

- Boiling point elevation

ΔT = Kb . m . i

- Freezing point depressión

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

3 years ago

A student, wearing chemical safety goggles and a lab apron, is to perform a laboratory test to determine the pH value of two dif

Setler79 [48]

They should probably wear gloves. A pH of 2.0 is a very strong acid, and it can easily irritate their hands. They should also remember to NEVER touch their face, mouth, etc. and to keep all chemicals on the laboratory table.

6 0

3 years ago

A sample of gas occupies a volume of 350.0 mL at 840mm Hg and 33°C. Determine the volume of this sample at 600 mm By and 52°C

insens350 [35]

Answer:

V₂ = 520.42 mL

Explanation:

Given data:

Initial volume = 350.0 mL

Initial pressure = 840 mmHg

Initial temperature = 33°C (33 +273 = 306 K)

Final temperature = 52°C (52+273 = 325 K)

Final volume = ?

Final pressure = 600 mmHg

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 840 mmHg × 350.0 mL × 325 K / 306 K × 600 mmHg

V₂ = 95550000 mmHg.mL.K /183600 K.mmHg

V₂ = 520.42 mL

7 0

2 years ago