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xxMikexx [17]
3 years ago
5

How much energy is required to melt a 500. gram block of iron? The heat of vaporization is 6090 J/g and the heat of fusion is 24

7 J/
g. 3.05x106 J 1.23x105 J 6.09x103 J 2.47x102 J
Chemistry
2 answers:
N76 [4]3 years ago
6 0

Give them brainliest ^^^

galben [10]3 years ago
3 0

The mass of iron block is 500 g. The amount of energy required to melt the iron block needs to be calculated. Melting means conversion of solid to liquid thus, heat of fusion is used which is 247 J/g.

From heat of fusion, 247 J of energy is released by melting 1 g of iron block. Thus, the amount of heat released by melting 500 g of iron rod will be:

H= 247 J/g× 500 g=1.23×10^{5}

Hence, option B is correct.

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If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank th
Svet_ta [14]

Answer:

F = 774146.534\,N

Explanation:

The pressure at the bottom of the tank is:

P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )

P_{bottom} = 73581.921\,Pa

The force exerted on the circular bottom is:

F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}

F = 774146.534\,N

4 0
3 years ago
Data was collected by students in an acid base titration lab. They used 1.63 M Ca(OH)2(AQ)
jenyasd209 [6]

Answer:

3.8 M

Explanation:

Volume of acid used VA= 57.0 - 37.5 = 19.5 ml

Volume of base used VB= 67.8 - 45.0 = 22.8 ml

Equation of the reaction

2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)

Number of moles of acid NA= 2

Number of moles of base NB= 1

Concentration of acid CA= ???

Concentration of base CB= 1.63 M

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA= CBVBNA/VANB

CA= 1.63 × 22.8 × 2/ 19.5 × 1

CA= 3.8 M

HENCE THE MOLARITY OF THE ACID IS 3.8 M.

6 0
3 years ago
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Answer:F

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3 years ago
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At 450 K the rate constant is 15.4 atm-1s-1. How much time (in s) is needed for NOCl originally at a partial pressure of 56 torr
Gwar [14]

This is an incomplete question, here is a complete question.

Consider the second-order decomposition of nitroysl chloride:

2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)

At 450 K the rate constant is 15.4 atm⁻¹s⁻¹. How much time (in s) is needed for NOCl originally at a partial pressure of 53 torr to decay to 10.6 torr?

Answer : The time needed for NOCl is, 2.52 seconds.

Explanation :  Given,

Rate constant = 15.4atm^{-1}s^{-1}

Initial partial pressure of NOCl = 56 torr = 0.0737 atm

final partial pressure of NOCl = 14.5 torr = 0.0191 atm

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant

t = time

[A_t] = concentration at time 't'

[A_o] = initial concentration

As we know that,

PV=nRT\text{ or }PV=CRT

Thus, the expression of second order kinetics will be:

kt=RT\times \frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}

\frac{k}{RT}t=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}

As, k'=\frac{k}{RT}

So, k't=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}         ............(1)

Now put all the given values in the above expression 1, we get:

(15.4atm^{-1}s^{-1})\times t=\frac{1}{0.0191atm}-\frac{1}{0.0737atm}

t=2.52s

Therefore, the time needed for NOCl is, 2.52 seconds.

5 0
3 years ago
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