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aivan3 [116]
3 years ago
5

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu

re is held constant
Chemistry
1 answer:
Arisa [49]3 years ago
8 0

Answer:

\boxed {\boxed {\sf P_2= 9 \ atm}}

Explanation:

Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

P_1V_1=P_2V_2

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

3.0 \ atm * 150 \ mL = P_2V_2

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

3.0 \ atm *150 \ mL= P_2 * 50 \ mL

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

450 \ atm*mL= P_2 * 50 \ mL

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.

\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}

\frac{450 \ atm*mL}{50 \ mL}= P_2

The units of milliliters will cancel.

\frac{450 \ atm}{50 }= P_2

9 \ atm =P_2

The new pressure is <u>9 atmospheres.</u>

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if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro
tensa zangetsu [6.8K]
<h3>Answers:</h3>

             1) 2 Units of Ozone

             2)  3 Units of Ozone

              3)  9 Units of Ozone

<h3>Solution:</h3>

1)  From 6 Oxygen Particles;

As given,

                          3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      6 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

                           3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      9 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  3 Units of Ozone

3)  From 27 Oxygen Particles;

As given,

                             3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      27 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  9 Units of Ozone

3 0
3 years ago
3.95 g of sugar (C6H12O6) is dissolved in water to make 158 mL of solution. Find the molarity.
12345 [234]

Answer:

[C₆H₁₂O₆] = 0.139 M

Explanation:

Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.

We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.

For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)

(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL

We determine moles, we convert them to mmoles, we divide by mL

M = 0.139 M

Moles = 3.95 g . 1mol / 180g → 0.0219 mol

We convert mL to L → 158 mL . 1L/1000mL = 0.158L

M = 0.0219 mol / 0.158L = 0.139 M

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Answer:

\Delta G=-541.4kJ/mol

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

H_2+F_2\rightarrow 2HF

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol

Regards!

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