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victus00 [196]
3 years ago
13

Why is a thermos considered a (mostly) closed system? Question 3 options: different types of liquids stay hot thermoses come in

different shapes and sizes different types of hot or cold items can be stored in a thermos matter and energy cannot enter or leave the system when the thermos lid is on tightly
Chemistry
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

Different types of hot or cold items can be stored in a thermos and power cannot enter or exit the system when the thermos lid is tightly closed

Explanation:

Closed systems are those that do not interact or do not exchange energy with the environment that surrounds them, that is why internal temperatures and conditions are maintained.

The human body is an open system, that is, it would be the opposite of the thermos since we constantly exchange energy with the environment through sweating, emission of gases, urine, feces, and the ingestion of food.

Thermoses are systems specially created to maintain a medium, it will be maintained if its lid is hermetically closed to prevent heat leakage or entry in situations of cold fluids.

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Looking at the following equation, what coefficient should be placed in front of the H2? Fe + H2SO4 --> Fe2(SO4)3 + H2
Hatshy [7]

Answer:

3 (three)

Explanation:

2 Fe + 3H2SO4 = Fe2(SO4)3 + 3 H2 (basically just balance both sides)

6 0
3 years ago
Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital
Thepotemich [5.8K]

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a <u>completely filled 5f, 6d, 7s and 7p orbitals</u>.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the <u>number of electrons</u> = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = <u>118</u>.

<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

7 0
3 years ago
What was the event that triggered Nicole’s deterioration
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Answer: A severe viral respiratory infection

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3 years ago
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For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati
Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

-N_A A = \frac{dm}{dt}

-N_A A = \frac{dVp}{dt}

-N_A A = \frac{dV}{dt}p

-N_A A = \frac{dhA}{dt}

-N_A A = \frac{dh}{dt} \times Ap

From the last step, area cancels out and thus leaves :

-N_A  = \frac{dh}{dt} \times p

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9}  \times (3 \times 10^{26}-C_{A2})=0.000246$$=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

3 0
3 years ago
Microwave ovens heat food by creating microwave electromagnetic radiation that is absorbed by water molecules in the food.
Nikolay [14]

Explanation:

It is known that relation between heat energy and temperature change is as follows.

             Q = mC \Delta T

where, q = heat energy

            m = mass of the solution or solvent

     \Delta T = change in temperature

Hence, putting the given values into the above formula as follows.

            Q = mC \Delta T

                = 110.0 g \times \frac{315 J}{75^{o}C} \times (100 - 25)^{o}C

                = 34650 J

Now, calculate the energy of photons at wavelength 3.02 mm as follows.

             E = \frac{hc}{\lambda}

where,     h = planks constant = 6.626 \times 10^{-34} Js

                c = velocity of light = 3 \times 10^{8} m/s

            \lambda = wavelength = 3.02 mm = 3.02 \times 10^{-3} m   (as 1 m = 1000 mm)

Therefore, putting the given values into the above formula as follows.

              E = \frac{hc}{\lambda}

                 = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{3.02 \times 10^{-3} m}

                = 6.58 \times 10^{-23} J

Now, the number of photons required of energy 6.58 \times 10^{-23} J/photon for the total energy of 34650 J as follows.

                E_{total} = n \times E

                     n = \frac{E_{total}}{E}

                        = \frac{34650 J}{6.58 \times 10^{-23}}

                        = 5265.95 \times 10^{23} photons

or,                     = 5.265 \times 10^{26} photons

Thus, we can conclude that the minimum number of photons present are 5.265 \times 10^{26}.

7 0
2 years ago
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