For the chemical reaction C2H6 --> C2H4 + H2 + 137kJ, the chemical energy of the

I am so lost, and suggestions to how I can solve this?

Ad libitum [116K]

<h3>Answer:</h3>

165.078 g

<h3>Explanation:</h3>

<u>We are given;</u>

Specific heat capacity of nickel as 0.446 J/g°C
Initial temperature of nickel is 255.5 °C
Mass of water in the calorimeter is 250.0 g
Initial temperature of water is 20° C
Final temperature of the mixture is 35.5°C

Specific heat capacity of water is 4.18 J/g°C

We are required to calculate the mass of nickel sample;

<h3>Step 1: Calculate the amount of heat absorbed by water.</h3>

We know that quantity of heat absorbed, Q = m × c × ΔT

Change in temperature, ΔT = 35.5°C - 20°C

= 15.5 °C

Therefore, Q = 250 g × 4.1 J/g°C × 15.5° C

= 16,197.5 Joules

<h3>Step 2: Calculate the amount of heat released by nickel sample</h3>

Assuming the mass of nickel sample is m

Then, Heat released, Q = m × c × ΔT

Change in temperature, ΔT = 255.5 °C - 35.5 ° C

= 220 °C

Q = m × 0.446 J/g°C × 220° C

= 98.12m joules

<h3>Step 3: Determine the mass of nickel sample</h3>

We know that the amount of heat absorbed is equivalent to amount of heat released.

That is, Quantity of heat absorbed = Quantity of heat released

Therefore;

98.12 m joules = 16,197.5 Joules

m = 165.078 g

Thus, the mass of nickel sample is 165.078 g

8 0

4 years ago

A 68-kg man whose average body temperature is 39°C drinks 1 L of cold water at 3°C in an effort to cool down. Taking the average

Dmitry [639]

Answer:

<h3>The man's average body temp. will fall by 0.6°C to 38.4°C</h3>

Explanation:

The enthalpy (heat content) of the water, using a datum of 0°C, is

Hw = Mw kg x Cp,w (specific heat capacity kJ/kg °C) x Tw °C

= 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ

Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ

So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.

But Hman (after drink) has mass 68 + 1 = 69 kg

Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C

So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew

Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C

8 0

3 years ago

The process which plants use to make food is called:

riadik2000 [5.3K]

The answer is Photosynthesis!

5 0

4 years ago

Read 2 more answers

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

gogolik [260]

<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.

<u>Explanation:</u>

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1): $H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)$

(2): $H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)$

(3): $H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)$

(4): $H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)$

Calculating the heat absorbed for the process having the same temperature:

$q=m\times \Delta H_{(f , v)}$ ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and $\Delta H_{(f , v)}$ is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

$q=m\times C_{s,l}\times (T_2-T_1)$ ......(ii)

where,

$C_{s,l}$ = specific heat of solid or liquid

$T_2\text{ and }T_1$ are final and initial temperatures respectively

<u>For process 1:</u>

We are given:

$m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC$

Putting values in equation (i), we get:

$q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J$

<u>For process 2:</u>

We are given:

$m=46g\\\Delta H_{fusion}=334J/g$

Putting values in equation (i), we get:

$q_2=46g\times 334J/g\\\\q_2=15364J$

<u>For process 3:</u>

We are given:

$m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC$

Putting values in equation (i), we get:

$q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J$

<u>For process 4:</u>

We are given:

$m=46g\\\Delta H_{vap}=2260J/g$

Putting values in equation (i), we get:

$q_4=46g\times 2260J/g\\\\q_4=103960J$

Calculating the total amount of heat released:

$Q=q_1+q_2+q_3+q_4$

$Q=[(2424.2)+(15364)+(19255.6)+(103960)]$

$Q=141003.8J=141.004kJ$ (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

7 0

3 years ago

What would be the effect on the calculated ratio (mole of copper:mole of iron) if:

kotegsom [21]

Explanation:

Since the ratio is Cu/Fe, if some Fe were lost due to spillage, the Cu/Fe ratio would INCREASE because Fe would be lower.

5 0

3 years ago