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Artemon [7]
3 years ago
13

What is the mass, in amu, of an atom with one proton and two neutrons?

Chemistry
1 answer:
Neko [114]3 years ago
4 0

mass of proton mp=1.672621898e-27 kg and 1 amu=1.66053904e-27kg so it’s just divition (1.672621898e-27 / 1.66053904e-27) =1.007276 amu

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What is the solubility of salt in water at room temperature
rewona [7]
<h3><u>Answer;</u></h3>

357 mg/mL

<h3><u>Explanation;</u></h3>
  • Solubility is defined to be the maximum amount of solute that will dissolve in a given amount of solvent at a specific temperature. The solubility of a salt is one of many physical properties that depend on temperature.
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5 0
3 years ago
PLEASE HELPP ASAP!!!!!What is the equation for density A. D=M/V B. D=V/M C. D= M/V2 D. D=1/2 MV2
mestny [16]

The correct answer is option A.

D = M/V

The density of a substance is the ratio of its mass to its volume.

Density = mass / volume

or D = M/V

The unit of density is gram per milliliter or g/ml, when mass is expressed in gram or g and the volume is expressed in milliliter ml.

If we know the mass and volume of a substance we can calculate its density using the formula for density.


8 0
3 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
3 years ago
In Period 2, as the elements are considered from left to
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Answer:

it will option B ,hope it helps

5 0
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Montano1993 [528]

Answer:

Why is copper used for most electrical wiring? All metals have some amount of resistivity to electrical currents, which is why they require a power source to push the current through. The lower the level of resistivity, the more electrical conductivity a metal has

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