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geniusboy [140]
4 years ago
5

If the temperature of a gas is raised from 30°C to 60°C, what happens to the pressure?

Chemistry
1 answer:
Fofino [41]4 years ago
5 0
<h3>Answer:</h3>

The pressure increases by 10% of the original pressure

Thus the new pressure is 1.1 times the original pressure.

<h3>Explanation:</h3>

We are given;

  • Initial temperature as 30°C, but K = °C + 273.15
  • Thus, Initial temperature, T1 =303.15 K
  • Final temperature, T2 is 333.15 K

We are required to state what happens to the pressure;

  • We are going to base our arguments to Pressure law;
  • According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume
  • That is; P α T
  • Therefore, at varying pressure and temperature

\frac{P1}{T1}=\frac{P2}{T2}

Assuming the initial pressure, P1 is P

Rearranging the formula;

[tex]P2=\frac{P1T2}{T1}[/tex]

P2=\frac{(P)(333.15K)}{303.15K}

     P2 = 1.099P

                 = 1.10 P

The new pressure becomes 1.10P

This means the pressure has increased by 10%

We can conclude that, the new pressure will be 1.1 times the original pressure.

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During a synthesis reaction, 2.4 grams of magnesium reacted with 8.0 grams of oxygen. What is the maximum amount of magnesium ox
natali 33 [55]

Answer: The maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.

Explanation : Given,

Mass of Mg = 2.4 g

Mass of O_2 = 8.0 g

Molar mass of Mg = 24.3 g/mol

Molar mass of O_2 = 31.9 g/mol

First we have to calculate the moles of Mg and O_2.

\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}

\text{Moles of }Mg=\frac{2.4g}{24.3g/mol}=0.099mol

and,

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}

\text{Moles of }O_2=\frac{8.0g}{31.9g/mol}=0.25mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

2Mg+O_2\rightarrow 2MgO

From the balanced reaction we conclude that

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

So, 0.099 moles of Mg react with \frac{0.099}{2}=0.049 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 Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of MgO

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.099 mole of Mg react to give 0.099 mole of MgO

Now we have to calculate the mass of MgO

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

Molar mass of MgO = 40.3 g/mole

\text{ Mass of }MgO=(0.099moles)\times (40.3g/mole)=3.9g

Therefore, the maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.

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28. Scott travels north 5 miles and then goes south 3 miles before coming back due north for an additional 2 miles. What is his
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Answer:

4 miles north

Explanation:

Initially Scott travels 5 miles north at this point his displacement is 5 miles toward north

then he goes 3 miles south. which makes his displacement 2 miles (5 - 3 = 2) towards north as north and south are opposite directions.

finally he travels 2 miles due north.so his final displacement is

2 miles + 2 miles = 4 miles towards north direction

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The amount of energy released or absorbed as heat during a chemical reaction is called ___________.
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