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inessss [21]
3 years ago
8

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

ing the specific heat of the bucket, calculate the final temperature of the two compounds once thermal equilibrium is reached
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
7 0
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

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I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
3 0
3 years ago
Which of the following compounds contains both ionic and covalent bonds? KOH N2O5 CH3OH Na2O
dusya [7]

Answer: a.KOH

Potassium hydroxide is an ionic compound where the K+ is the cation and OH−is the anion. At the same time, the compound also contains a covalent bond since the anion, OH−is formed from electron sharing between the O and H atoms.

Hope this helps........ Stay safe and have a Merry Christmas! :D

Explanation:

4 0
3 years ago
A baby weighs 7 pounds, 4 ounces at birth and 7 pounds, I ounce at discharge. What percent of weight did the baby lose
Aloiza [94]

Answer:

2.6%

Explanation:

As, 1 ounce (oz) =  0.0625 pounds (lb)

Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb

Since, 1 oz =  0.0625 lb

⇒ 4 oz = 4×0.0625 = 0.25 lb

Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb

The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.

Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>

Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.

Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>

6 0
3 years ago
How many grams of sulfur are needed to react completely with 246 g of mercury to form hgs?
iren [92.7K]

The reaction between mercury (Hg) and sulfur (S) to form HgS is:

Hg + S ------------- HgS

Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS

The given mass of Hg = 246 g

Atomic mass of Hg = 200.59 g/mol

# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles

Based on the reaction stoichiometry,

# moles of S that would react = 1.226 moles

Atomic mass of S = 32 g/mol

Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g

39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS

5 0
3 years ago
Read 2 more answers
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
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