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Elena-2011 [213]
3 years ago
7

Which is an electron dot diagram of an ionic compound?

Physics
1 answer:
yuradex [85]3 years ago
8 0
C.NaCl as ionic compounds are made of a metal and a non-metal and they form oppositely charged ions hence the name ionic bonding which forms ionic compounds.The other are molecules with covalent bonding.
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it is D exo means realease

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If Michael Jordan has a vertical leap of 1.29 meters, what is his take-off speed and his hang time (total time to move upwards t
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(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>
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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
2 years ago
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

Q = ms \Delta T

Q = 1*s*(250 - 20)

Q = 1*s*230

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

Q = ms'\Delta T

1*s*230 = 1* s' * \Delta T

now we have

\Delta T = (\frac{s}{s'})*230

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

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The independent variable:
ValentinkaMS [17]

Answer:

A. Is the one that the experimenter manipulates directly

Explanation:

The independent variable is the one that is manipulated during an experiment by the experimenter.

The dependent variable is the one that is effected by the independent variable in an experiment.

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