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Lubov Fominskaja [6]
3 years ago
15

A gas at a pressure of 360 torrs and a volume of 750 ml. If the pressure becomes 1.5 atm what is the new volume?

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

V₂ = 236.84 mL

Explanation:

The relation between pressure and volume is inverse.

We can write it as follows :

P_1V_1=P_2V_2

We have,

P₁ = 360 torrs, V₁ = 750 mL, P₂ = 1.5 atm = 1140 torr.

So,

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{360\times 750}{1140}\\\\V_2=236.84\ mL

So, the new volume of the gas is 236.84 mL.

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Answer:

Choice A: Approximately 5.23 \times 10^{-29} joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the \rm 5.81\times 10^{-29} kilograms of mass.

E = m \cdot c^{2},

where

  • E stands for energy,
  • m stands for mass, and
  • c is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of c?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}.

Apply the mass-energy equivalence:

\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}.

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

\rm 1\; J = 1\; N \times 1\; m.

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}.

In other words, the mass defect here is also \rm 5.23\times 10^{-12}\; J.

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In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca
Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

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Explanation:

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So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

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