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borishaifa [10]

3 years ago

9

Jenny wants to test the electrical conductivity of two substances dissolved in water. She is preparing the containers for the ex

periment. Which factor is MOST important for her to control?

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

Volume of the solutions

This is the most important factor for her to control.

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I have 345mL of a 1.5M NaCl solution. If I boil the water until the volume of the solution is 250mL, what will the molarity of t

Karo-lina-s [1.5K]

Answer is
C. 2.07 M

For explanation

M1V1 = M2V2

M2 = (M1V1)/V2

M2 = (1.5M x 345ml) / 250 ml

:. M2 = 2.07 M

8 0

3 years ago

Laney left a nail in water for two weeks, and the nail became rusted from sitting in the water.

Simora [160]

B) A chemical change because the nail reacts with water/oxygen to create rust (a type of oxide)

5 0

3 years ago

Exploring the Second Law

san4es73 [151]

Answer:

(a) The coefficient of performance of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when both exchange energy by heat transfer with the same two reservoirs.

Explanation:

According to the Kelvin–Planck statement of the second law of thermodynamics ,it is not possible to construct a device which operates in cycle and does not produce effect on the environment than the production of work.

We know that

Coefficient of performance is the ratio of desired effect to the work input in a cycle.

Given all option is correct but most appropriate option is a.

So the option a is correct

(a) The coefficient of performance of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when both exchange energy by heat transfer with the same two reservoirs.

5 0

3 years ago

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0

3 years ago

How many mols of H2SO4 is in 0.80 L of 1.40 M H2SO4?

Hitman42 [59]

I think it’s 6 yw. Sorry if I’m wrong

7 0

2 years ago