All categories

3 years ago

1200 scientific notation

Chemistry

2 answers:

Mnenie [13.5K]3 years ago

7 0

Answer:

1.2 x 10 the third power

Explanation:

Whitepunk [10]3 years ago

3 0

Answer:

1.2 x $10^{3}$

You might be interested in

HELP TIMED !!!!

Troyanec [42]

Hello there what is the question?

6 0

4 years ago

Read 2 more answers

Choose a starting point in the water cycle and describe the process you would go through to move through the entire cycle.

Bogdan [553]

The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation. ... The cycling of water in and out of the atmosphere is a significant aspect of the weather patterns on Earth.

4 0

3 years ago

What type of changes indicate a physical change?

Illusion [34]

vaporisation, Freezing, Condensation

6 0

4 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

A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti

lilavasa [31]

c = 1.14 mol/L; b = 1.03 mol/kg

Molar concentration

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

c = 1.14 mol/1 L = 1.14 mol/L

Molal concentration

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

b = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0

3 years ago

Read 2 more answers