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3 years ago

I WILL GIVE OUT BRAINLIEST IF CORRECT

Chemistry

2 answers:

aivan3 [116]3 years ago

8 0

The answer is D.

The reason that I say the answer is D is because it includes information about how pumping the air into the ball is a closed system meaning that the Kinetic energy is going to collide due to being forced into the ball, and once the space is full, then it has collided.

_______________________________________________________

Hope this helps.

-Northstar

ololo11 [35]3 years ago

7 0

Answer:

the answer is D my friend

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In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form

Mkey [24]

The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein.

E = Δmc²
where
E is the energy gained or released during the reaction
c is the speed of light equal to 3×10⁸ m/s
Δm is the mass change

(1.715×10³ kJ)(1,000 J/1 kJ) = Δm(3×10⁸ m/s)²
Δm = 1.91×10⁻¹¹ kg

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3 years ago

Which is NOT a component of galaxies? * F. Universe G. Stars H. Dust J. Planets

Afina-wow [57]

Answer:

F. Universe

Explanation:

Universe and Galaxies are two different things.

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3 years ago

What is the percent by mass of Fluorine in Nitrogen triflouride NF3?

Kipish [7]

Answer:

The answer to your question is 80.3%

Explanation:

Data

Percent by mass of F

molecules NF₃

Process

1.- Calculate the molar mass of nitrogen trifluoride

molar mass = (1 x 14) + (19 x 3)

= 14 + 57

= 71 g

2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

71 g of NF₃ ------------------ 100%

57 g of F ------------------- x

x = (57 x 100)/71

x = 5700 / 71

x = 80.3%

3.- Conclusion

Fluorine is 80.3% by mass of the molecule NF₃

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3 years ago

What observations can you make about this group of people? Name one inference you can make from your observations.

Vlad1618 [11]

Answer:

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7 0

3 years ago

Read 2 more answers

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago