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3 years ago

In the morning a student puts a warm soda in a ice filled cooler.

1 answer:

4 0

basically all of the molecules in the can of soda started to stop moving due to the compression and change of temperature. the ice had frozen the molecules and which then made the can of soda nice and cold lol

You might be interested in

How many grams are there in 2.4 moles of CaCO3?

Montano1993 [528]

I believe it’s 5.7 moles

5 0

2 years ago

What mass of fes is formed if 9.42 g of fe reacts with 68.0 g of s8? 8fe(s) + s8(s) â†’ 8fes(s)?

allochka39001 [22]

From the equation, we see that the molar ratio of Fe : S required is:

8 : 1

The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265

The molar ratio is:

1 : 1.6

Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of

Fe : FeS is
1 : 1

So 0.168 moles of FeS will form. The mass of FeS will be:

Mass = 0.168 * (56 + 32)
Mass = 14.78 grams

14.78 grams of FeS will be formed.

4 0

3 years ago

In the formula

faust18 [17]

Answer: remove a H2 molecule from the left side of the equation

Explanation:

6 0

2 years ago

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

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

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

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

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

2 years ago

A system receives 575 ) of heat and delivers 425 ) of work. Calculate the change in the internal energy. AE, of the system.

vfiekz [6]

Answer:

ΔE = 150 J

Explanation:

From first law of thermodynamics, we know that;

ΔE = q + w

Where;

ΔE is change in internal energy

q is total amount of heat energy going in or coming out

w is total amount of work expended or received

From the question, the system receives 575 J of heat. Thus, q = +575 J

Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.

Thus;

ΔE = 575 + (-425)

ΔE = 575 - 425

ΔE = 150 J

7 0

3 years ago