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Nat2105 [25]
3 years ago
14

a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa

ctor of 2 what will happen to the pressure? you may assume the temperature remains at 35 c
Chemistry
2 answers:
lesya692 [45]3 years ago
6 0

Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

Where:

V₁ = Initial volume

V₂ = Final volume = V₁/2

P₁ = Initial pressure = 832 torr

P₂ = Final pressure  = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂  = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

Nonamiya [84]3 years ago
4 0

Answer:

The pressure will increase by a factor of 2 and is now 1664 torr

Explanation:

The question says for us to assume that temperature is constant. Now, since we are given pressure and volumw, we will use Boyle's law which is a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature i.e V ∝ 1/P

Thus, PV = K

Where K is a constant.

So,

P₁•V₁ = P₂•V₂ = k

Where:

P₁ = Initial pressure

V₁ = Initial volume

P₂ = Final pressure

V₂ = Final volume = V₁/2

From the question, P₁ = 832 torr ; we are told that volume is decreased by 2,thus, V₂ = V₁/2

Now, we want to find out what will happen to the pressure P₂.

Let's make P₂ the subject;

P₁•V₁ = P₂•V₂

Thus, P₂ = (P₁•V₁)/V₂

Plugging in the relevant values to obtain ;

P₂  = (P₁•2V₁)/V₁

V₁ will cancel out and we have;

P₂  = 2P₁

Now, we are given that P₁ = 832 torr, Thus,

P₂ = 2P₁ = 2× 832 torr = 1664 torr

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Serga [27]

Answer:

Position D

Explanation:

One fourth of the Moon will be visible from Earth when the Moon is in position D. The reason why this is the right position is that the Moon will be in a position where the majority of its surface that is lighted by the Sun is facing the Sun. On the other hand, the angle is just right that about one quarter of the lighted part of the Moon is also facing the Earth. This will result in a perception from the Earth's view point that the Moon is lighted at one quarter of its surface.

6 0
3 years ago
determine the molecular formula of the compound with an empirical formula of CH and a molar mass of 78.110g/mol
Vlada [557]

Answer:

C_{6} H_{6}

Explanation:

First, find the mass of empirical formula, CH. 12.01 g/mol is for carbon, and 1.008 g/mol is for hydrogen. 12.01+1.008=13.018 G/mol CH. Divide 78.110 G/mol by 13.018 g/mol. You get approximately 6. Multiply that by the subscript of each element. 6(CH)=

C_{6} H_{6}

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3 years ago
Chris noticed a high-pressure system in the weather forecast for Monday
Alchen [17]

Answer:

A. Clear and Sunny

Explanation:

The answer would be A. Clear and Sunny. A high pressure system occurs where the air mass above the Earth is denser than in surrounding areas, and therefore exerts a higher force or pressure. They usually happen with li9ght winds. Using process of elimination, it could not be D. Warm and Stormy, C. Cloudy and Rainy, or B. Cold and Stormy because all have wet climates.

5 0
2 years ago
How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

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The answer to your question is a 12

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