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Hitman42 [59]
3 years ago
10

The molar enthalpy of fusion for water is 6.008 kj/mol. what quantity of energy is released when 253g of liquid water freezes? (

molar mass of water is 18.02 g/mol)
Physics
1 answer:
e-lub [12.9K]3 years ago
6 0
During freezing, energy is released by the mass of water without change in temperature. Such energy will also be required if the same mass of water has to be melted.

Then,

Number of moles = mass/molar mass = 253/18.02 =14.04 moles

Energy released = moles*molar enthalpy of fusion = 14.04*6.008 = 84.35 kJ
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HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
How many significant figures does 0.00340 have?
True [87]
Greetings!

"<span>How many significant figures does 0.00340 have?"...

A significant figure is:
-A non zero number
-A zero in between non zero numbers
-Trailing zeros to the right of the decimal point

Therefore, this number has three significant figures:
3,4,0

Hope this helps.
-Benjamin</span>
4 0
3 years ago
Read 2 more answers
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

s=720.3\ m

We need to calculate the velocity

Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

H'=2881.2\ m

Hence, The maximum height is 2881.2 m.

4 0
3 years ago
Given the scope of Kayla’s education, training, and years of experience as a CMA, would this “favor” fall within the AAMA guidel
Cerrena [4.2K]

Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.

From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.

Even though the favor that was asked by Dr. Hsu was prohibited by AAMA, it should be noted that the final part of the favor about faxing the report to the internist would fall within AAMA guidelines.

In conclusion, the best way that Kayla can respond to Dr. Hsu is to decline doing the favor.

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6 0
2 years ago
Can someone help me?!!!!!
ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

(-3\cdot (-6), -2 \cdot (-6))=(18,12)

The magnitude is given by Pythagorean's theorem:

m=\sqrt{18^2+12^2}=21.6

and the direction is given by the arctan of the ratio between the y-component and the x-component:

\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}

3 0
3 years ago
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