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Scrat [10]
2 years ago
5

How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper

ature of 75 C
Chemistry
1 answer:
satela [25.4K]2 years ago
5 0

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

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Consider a general reaction
choli [55]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

7 0
3 years ago
If your density was supposed to be 2.3 g/mL, but you calculated yours to be 2.1 g/mL, what is your percent error?
Reptile [31]

Answer: 0.08695652

Explanation:

You would do the answer you got subtracting from the expected answer over your expected answer

6 0
3 years ago
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
2 years ago
Define the term inertia
n200080 [17]

Answer:

Explanation:

Enertia is an integral part of Newton's first law of motion.

It is the tendency of an object to <u>stay at rest</u> or <u>to continue moving</u> until and unless <u>any external unbalanced force</u>, (like, applied force or force of tension or frictional force ) is applied to either move it from rest or change its speed(in other words, accelerate it!!).

Example below, is of ball at rest (fig1) and if this ball is moving straight on a frictionless surface(like ice) it will keep moving!! until, we push it or pull it.

5 0
3 years ago
Fill in the chart to describe and give examples of physical changes.
frosja888 [35]

Answer:

here are some examples of physical change!!!

Explanation:

-An ice cube melting into water in your drink.

-Freezing water to make ice cubes.

-Boiling water evaporating.

-Hot shower water turning to steam.

-Steam from the shower condensing on a mirror.

8 0
2 years ago
Read 2 more answers
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