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Mandarinka [93]
2 years ago
6

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
Kitty [74]2 years ago
6 0

Answer:

Q = 282,000 J

Explanation:

Given that,

The mass of liquid water, m = 125 g

Temperature, T = 100°C

The latent heat of vaporization, Hv = 2258 J/g.

We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :

Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J

or

Q = 282,000 J

So, the required heat is 282,000 J .

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Help with this question please.
mr Goodwill [35]

There are 4 hydrogens on the right side (2\mathrm H_2=4\mathrm H), and 2 hydrogens on the left per molecule of \mathrm H_2. To get the same number of hydrogens on both sides, the coefficient should be 2.

(Then the number of oxygens will be consistent, since 2\mathrm H_2\mathrm O contributes 2 oxygens, and so does \mathrm O_2.)

5 0
3 years ago
How should the student change the circuit to give negative values for current and
natulia [17]

Answer:

changing the polarity or direction of the battery changes the sign of the voltage and the current

Explanation:

The sign of current and voltage are due to established conventions.

The way that a DC circuit with negative current values ​​is by changing the polarity of the power source or by inverting the battery, this creates that the electrons move in the opposite direction

Changing the battery also changes the direction of the power difference, since the potential from positive to negative, in most cases negative is assigned a potential of zero volts

In summary, changing the polarity or direction of the battery changes the sign of the voltage and the current

3 0
2 years ago
When a cold air mass and a warm air mass meet but are at a standstill, the boundary is called.
Sphinxa [80]

Answer:

Ur answer is Stationary Front

Explanation:

Stationary Front is when a cold air mass and a warm air mass but are at a standstill the boundary is called Stationary Front.

3 0
2 years ago
An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.
adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

\Delta x\times \Delta p=\frac{h}{4\pi} ...........(1)

where,

\Delta x = uncertainty in position

\Delta p = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

p=m\times v

or,

\Delta p=m\times \Delta v      .......(2)

Equating 1 and 2, we get:

\Delta x\times m\times \Delta v=\frac{h}{4\pi}

\Delta v=\frac{h}{4\pi \Delta x\times m}

Given:

m = mass of electron = 9.11\times 10^{-31}kg

h = Planck's constant = 6.626\times 10^{-34}Js

radius of atom = 67.0pm=67.0\times 10^{-12}m     (1pm=10^{-12}m)

\Delta x = diameter of atom = 2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m

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

\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}

\Delta v=4.32\times 10^{5}m/s

The minimum uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

Now we have to calculate the percentage of the average speed.

Percentage of average speed = \frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100

Uncertainty in speed = 4.32\times 10^{5}m/s

Average speed = 1.3\times 10^{6}m/s

Percentage of average speed = \frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0
3 years ago
The magnitude of the charge of the electron is:
brilliants [131]

Answer:

a. Exactly the same as the magnitude of the charge of the proton.

Explanation:

The elementary charge (e) is the smallest electric charge that can exist in the universe. Any positive or negative electric charge can be expressed as a multiple of the elementary charge, since is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron (-1e).

3 0
3 years ago
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