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marysya [2.9K]
3 years ago
14

The temperature of a gas rose from 250K to 350K. At 350K, the volume of the gas was 3.0L. If the pressure did not change, what w

as the initial volume of the gas?
Please give an explanation and I will give you Brainliest :)
Chemistry
2 answers:
LenaWriter [7]3 years ago
7 0
Hey there,

So. . I believe is how you do it. I did 350 x 3.0 and it got me 1,050.
We always multiply it by when it come to the initial volume of the gas.

Hope this helps.

~Jurgen<span />
Oksi-84 [34.3K]3 years ago
7 0
Do 350,000 x 3.0 = 1,050,000 so you would do 250,000 x 3.0
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one kilogram of water (V1 = 1003 cm^3*kg-1) in a piston cylinder device at (25°C) and 1 bar is compressed in a mechanically reve
vova2212 [387]

Answer:

Q = -18118.5KJ

W = -18118.5KJ

∆U = 0

∆H = 0

∆S = -60.80KJ/KgK

Explanation:

W = RTln(P1/P2)

P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K

W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)

∆U = Cv(T2 - T1)

For an isothermal process, temperature is constant, so T2 = T1

∆U = Cv(T1 - T1) = Cv × 0 = 0

Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ

∆H = Cp(T2 - T1)

T2 = T1

∆H = Cp(T1 - T1) = Cp × 0 = 0

∆S = Q/T

Mass of water = 1kg

Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg

∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK

4 0
3 years ago
When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of O_2 = 8.00 g

Molar mass of O_2 = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and C is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0
3 years ago
Help plzzzzz ASAP!!!!!!!!
SIZIF [17.4K]

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

<h3>Further explanation</h3>

Given

Ionic charge

Required

The formula of binary ionic compounds

Solution

Ionic compounds consisting of cations (ions +) and anions (ions -)

Ionic compounds usually consist of metal cations and non-metal anions

Metal: cation, positively charged.

Nonmetal: negatively charged

The anion cation's charge is crossed

The ionic compounds :

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

4 0
3 years ago
Why does buck ministerfullerene act as a good lubricant
MakcuM [25]

Answer:

Its molecules are made up of 60 carbon atoms joined together by strong covalent bonds. Molecules of C 60 are spherical. There are weak intermolecular forces between molecules of buckminsterfullerene. These need little energy to overcome, so buckminsterfullerene is slippery and has a low melting point.

Explanation:

5 0
3 years ago
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