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Alex_Xolod [135]

3 years ago

6

HELP MEEEEEEE PLSSS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1 As distance increases between a rocket ship and the ground, gravitationa

l force remains the same decreases increases pushes less

1 answer:

Sladkaya [172]3 years ago

8 0

Answer:

decreases

Explanation:

Gravirational force is directly proportional to the mass and inversely proportional to the distance.(Newton's law of gravitation)

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What is the difference expressed in kilograms between the mass of proton and the mass of electon?

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Particle, Relative Mass<span>, Actual </span>Mass<span> (g) </span>Electron<span>- 1 , 9.11*10 power -28 g. ... Calculate: What is the diff </span>expressed in kg between the mass<span> of a </span>proton and the mass<span> ... Best Answer: </span>difference<span>in </span>mass<span> = [1.673 x 10^(-24) - 9.11 x ...</span>

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4 years ago

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4 years ago

What fundamental building block makes up all matter in the universe

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6 0

3 years ago

Which of the following represents the ground state electron configuration for the Mn3 ion?

Allisa [31]

Answer:

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$

Explanation:

The $Mn^{+3}$ has 22 electrons arround the nucleous. Its ground state configuration following the diagonal rule is:

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$

In term of the previous noble gas:

$[Ar] 4s^2 3d^2$

7 0

3 years ago

The normal boiling point of liquid ethyl acetate is 350 K. Assuming that its molar heat of vaporization is constant at 34.4 kJ/m

Roman55 [17]

Answer:

337.22 K

Explanation:

Given that:

P₁ = 1 atm

T₁ = 350 K

P₂ = 0.639 atm

T₂ = ??? (unknown)

R(rate constant) = 8.34 J k⁻¹ mol⁻¹

Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.

Clausius-Clapeyron equation can be written as:

$In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]$

Substituting our values given; we have:

$In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]$

$In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]$

$- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K$

$-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} *350T_2 =T_2-350$

$- 0.037889T_2=T_2-350$

$350 = 0.03789T_2+T_2$

$350 =1.03789T_2$

$T_2= \frac{350}{1.03789}$

$T_2 = 337.22K$

∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is <u>337.22</u> K.

5 0

3 years ago