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lisov135 [29]
3 years ago
9

If anyone is good at figuring out the amount of heat released in chemistry PLS HELP!!!!!

Chemistry
1 answer:
KATRIN_1 [288]3 years ago
3 0

Answer:

Option (3) 334J.

Explanation:

The following data were obtained from the question:

Mass (m) = 1g

Heat of fusion (ΔHf) = 334J/g

Heat (Q) =?

The heat released can be obtained as follow:

Q = m·ΔHf

Q = 1 x 334

Q = 334J.

Therefore, the heat released when 1g is converted to ice is 334J.

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I think it's A but it might be different from your school and mine
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Luda [366]
According to my calculations the correct Answer is C. +694. The answers A B and D Are incorrect. It cannot be A because I said it also cannot be B because C=H. The answer would not be D because balls in your jaws = +1192 And the answer will be CORRECT ANSWER C=+694
8 0
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Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

d= 50.23 g/cm³

8 0
3 years ago
Pls help me out!! lol
ella [17]

Answer:

i honestly dont know

Explanation:

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7 0
3 years ago
What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0
4 years ago
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