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3 years ago

Given the standard entropy values in the table, what is the value of º for the following reaction?

1 answer:

8 0

Hello.

The answer is <span>+313.766 J/mol·K
</span>
Use the coefficients of the reaction and sum the product entropies less the reactant entropies:

4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K

Have a nice day

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Match each state of matter to its description.

WARRIOR [948]

Answer:

Explanation:

solid would be a wall e.g

liquid would be water e.g

gas would be oxygen e.g

6 0

2 years ago

What is mean by kinetic energy

Vinvika [58]

The energy that an object has is moving

5 0

3 years ago

The period of a ripple on a lake is half a second. what is the frequency of the wave

Nataly_w [17]

Answer:

2 Hertz

Explanation:

<em>The frequency would be 2 Hertz.</em>

<u>The frequency of a wave is defined as the rate at which the particles of a medium vibrates when the wave is passed through it while the period of a wave is the time it takes the particles to make a complete cycle of vibration.</u>

The frequency of a wave is inversely related to its period and is defined by the following equation:

f = 1/t, where f is the frequency (in hertz) and t is the period (in seconds).

Hence, if the period of a ripple is 1/2 or 0.5 seconds, the frequency becomes;

f = 1/0.5 = 2 Hertz

7 0

3 years ago

Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11

otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

$2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O$

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = $\frac{2400 g}{890 g/mol}=2.6966 mol$

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;

$\frac{110}{2}\times 2.6966 mol=148.31 mol$ of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0

3 years ago

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

Diano4ka-milaya [45]

<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

7 0

3 years ago