All categories

3 years ago

Directions are in the picture

Chemistry

1 answer:

Mrrafil [7]3 years ago

6 0

Answer:

Question Clear

Explanation:

Make sure your question is clear.

You might be interested in

xxMikexx [17]

Answer:

its sunspots :)

Explanation:

can i pleasee have the brainiest tyty

for my rank

6 0

3 years ago

How many miles of N2 are in a flask with a volume of 250.0 mL at a pressure of 300.0 kPa and a temperature of 300.0 K?

Maslowich

As hot as my cooter lips

hope this helped :)

8 0

3 years ago

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

gogolik [260]

Answer: The amount of heat absorbed is 141.004 kJ.

Explanation:

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

For process 1:

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:

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

For process 2:

We are given:

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

Putting values in equation (i), we get:

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

For process 3:

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:

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

For process 4:

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

PLS HELP DECODE THIS

Nadya [2.5K]

The atomic number tells you the PROTONS

6 0

3 years ago

Convert 45 joules to heat calories

skad [1K]

Answer:

45 joules to calories= 10 (10.755258) calories

3 0

4 years ago