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2 years ago

When rubbing alcohol and water are at room temperature, which one boils first

Chemistry

2 answers:

Ugo [173]2 years ago

6 0

Answer:

rubbing alcohol

mylen [45]2 years ago

5 0

Answer:

alcohol

Explanation:

Alcohol has a lower boiling point

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Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is th

lubasha [3.4K]

Answer:

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

C₃H₈ + O₂ ⇒ CO₂ + H₂O

balanced chemical reaction

C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂O

Reactants Elements Products

3 C 3

8 H 8

10 O 10

b) Standard enthalpy

Propane -104.7 kJ/mol

Oxygen 0 kJ/mol

Carbon dioxide -393.5 kJ/mol

Water -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

4 0

3 years ago

What is the mass of 23.2 moles of Gallium atoms?

Effectus [21]

Explanation:

1 grams Gallium to mol = 0.01434 mol

10 grams Gallium to mol = 0.14342 mol

20 grams Gallium to mol = 0.28685 mol

30 grams Gallium to mol = 0.43027 mol

40 grams Gallium to mol = 0.5737 mol

50 grams Gallium to mol = 0.71712 mol

100 grams Gallium to mol = 1.43425 mol

200 grams Gallium to mol = 2.86849 mol

6 0

3 years ago

A round character is one-sided and does not show much of his/her emotions or conflicts.

My name is Ann [436]

Answer:

true

Explanation:

Round characters tend not to show a lot of their emotions or feelings.

7 0

3 years ago

In the following table, all the columns for the element beryllium are filled out correctly.

scoundrel [369]

That is false..........

6 0

4 years ago

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago