Urgent please please I really need help I will give you 60 points just please help

A) Devices that transfer kinetic energy have what for their source of power?

GarryVolchara [31]

Answer:

A) Devices that transfer kinetic energy have a source of power that is in motion

Kinetic energy is the energy in motion, as such, a device that transfers kinetic energy transfers the energy the power source has into other energy forms

B) Kerosene does not easily cold start like diesel which can burn after compression

C) The first law of thermodynamics states that energy is conserved and it can neither be created nor destroyed, but can be changed from one form to another.

Therefore, when energy is not available in a given location or body, it cannot be obtained from that body or location

Explanation:

4 0

4 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

. How many grams of water would require 4400 joules

Alja [10]

Answer:

The specific heat of water is 4.18 J/g C.

Explanation:

q

=

m

C

s

Δ

T

Never forget that!

2200

=

m

⋅

4.18

J

g

⋅

°

C

⋅

66

°

C

∴

m

≈

8.0

g

5 0

3 years ago

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0

3 years ago

The calculation of quantities in chemical equations are called...

Anni [7]

The calculation of quantities in chemical equations are called Stoichiometry. Stoichiometry is a branch of chemistry which deals with relative quantities of reactants and products in chemical reactions. The correct answer is 'Stoichoimetry'. I hope this helps you.

5 0

3 years ago