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9 days ago

A sample of neon gas at STP is allowed to expand into an evacuated vessel. What is the sign of work for this process?

Chemistry

1 answer:

Alekssandra [968]9 days ago

3 0

Answer:

The work done in this process will be considered Negative.

Explanation:

The energy transferred by the system to the environment is negative

Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.

In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.

Thus, the result for the work pertaining to this process will carry a Negative sign.

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Dicarbon monoxide, C2O, is found in dust clouds in space. Analysis of it shows that the sequence of atoms in this molecule is C–

castortr0y [923]

Greetings,

The number of lone pairs of electrons in a C2O molecule is...

Each Oxygen atom forms two bonds with Carbon.

I hope this was useful!

-Char

8 0

9 days ago

Read 2 more answers

A water tank can hold 1 m3 of water. When it’s empty, how much liters is needed to refill it?

Alekssandra [968]

Response: 1000

Rationale: because 5 cubic meters equals 5000 liters

3 0

5 days ago

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if

castortr0y [923]

Answer:

The correct options include choice 2, 3, and 6.

Explanation:

Density is identified as the mass of a substance per unit volume occupied by that substance.

$Density=\frac{Mass}{Volume}$

The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.

2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.

6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.

The same metals in both instances will yield consistent densities due to the fixed density of the metal.

7 0

3 days ago

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

eduard [944]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2. $5.6\cdot 10^{13}$ electrons per individual

Explanation:

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

$v_d=\frac{I}{neA}$

where

I stands for current

n represents the density of free electrons

$e=1.6\cdot 10^{-19}C$ indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

$A=\pi r^2$

where r denotes the wire's radius. Thus, the equation transforms to

$v_d=\frac{I}{ne\pi r^2}$

In this scenario, we have:

I = 8.0 A as the current

$8.5\cdot 10^{28} m^{-3}$ indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

So, the resulting drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

The entire length of the cord is

L = 3.00 m

And the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Consequently, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , thus the total electrons in this cord would be

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

Overall, the Earth's population rounds to 8 billion individuals, equating to

$N'=8\cdot 10^9$

Hence, the number of electrons distributed to each person is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

5 days ago

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

Tems11 [846]

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

7 0

14 hours ago