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Irina-Kira [14]

2 years ago

9

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1 answer:

mixer [17]2 years ago

3 0

Answer:

gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata

Explanation:

vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs

You might be interested in

Perform the following conversions:

ololo11 [35]

Answer:

a) $20^0C$ and $293K$

b) $437K$ and $327.2^0F$

c) $-273^0C$ and $459.44^0F$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ , $^0F$ and $K$

These units of temperature are inter convertible.

$t^0C=(t+273)K$

$t^oC=\frac{5}{9}\times (t^oF-32)$

$K-273=\frac{5}{9}\times (^oF-32)$

a) 68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into $^0C$ and $K$ by using conversion factor:

$t^oC=\frac{5}{9}\times (68^oF-32)$

$t=20^0C$

$20^0C=(20+273)K=293K$

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins and Fahrenheit

$164C=\frac{5}{9}\times (t^oF-32)$

$t^0F=327.2$

$164°C=(164+273)K=437 K$

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

$K-273=\frac{5}{9}\times (t^oF-32)$

$0-273=\frac{5}{9}\times (t^oF-32)$

$t=-459.4^0F$

$t^K=(0-273)^0C=-273^0C$

8 0

2 years ago

For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.

defon

1. Increasing
2. Increasing
3. Decreasing
4. Increasing
5. Decreasing

6 0

2 years ago

A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in

ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

First let's convert 25 Kg to g;

1 Kg = 1000 g

25 Kg = ?

$= \frac{25 × 1000}{1}$

= 25000 g

Substitute the values into the formula:

$p = \frac{25000 g}{18.3 ml}$

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is 1366.120 g/mL.

8 0

1 year ago

A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vine

QveST [7]

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:

Molarity = moles / volume

moles = Molarity * volume

0.4293 M * 39.27 mL = 16.86 mmol NaOH

<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M

4 0

2 years ago

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?$

Putting values in above equation, we get:

$1.0\times 3L=3.5\times V_2\\\\V_2=0.86$

Thus the squirrel's balloon will be 0.86 L

7 0

2 years ago