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SVEN [57.7K]
2 years ago
12

a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?

Chemistry
1 answer:
Zina [86]2 years ago
4 0

Answer:

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ =  70 KPa × 600 mL × 313 K / 293K ×150 mL  

P₂ = 13146000 KPa .mL. K /43950 K.mL

P₂ = 299.11 KPa

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c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produ
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<u>Answer:</u> 72 grams of water will be produced.

<u>Explanation:</u>

To calculate the number of moles, we use the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}    ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

\text{moles of propane}=\frac{44g}{44g/mol}=1mole

For the reaction of combustion reaction of propane, the equation follows:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = \frac{1}{1}\times 4=4moles of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

4mol=\frac{\text{Mass of water}}{18g/mol}

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6 0
3 years ago
How many grams of CO2 are in 3.6 mol of CO2?
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There are 158.4 grams of CO2 in 3.6 mol of CO2.

<h3>HOW TO CALCULATE MASS?</h3>

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;

mass of CO2 = no. of moles × molar mass

According to this question, there are 3.6 moles of CO2.

mass of CO2 = 3.6 moles × 44g/mol

mass of CO2 = 158.4g.

Therefore, there are 158.4 grams of CO2 in 3.6 mol of CO2

Learn more about mass at: brainly.com/question/15959704

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2 years ago
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Answer:

Reducing sugars are absent

Explanation:

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Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.

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3 years ago
49.9 ml of a 0.00292 m stock solution of a certain dye is ddiluted to 1.00 L. the diluted solution has an absorbance of 0.600. w
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Complete Question

49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye

Answer:

The  value is  \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

Explanation:

From the question we are told that

   The volume of the stock solution is  V_1   =  49.9 mL  =  0.0499 \  L  

   The concentration of the stock solution is  C_1  =  0.00292 \  M

   The volume of the diluted solution is  V_2 =  1.00 \  L

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Generally the from the titration equation we have that

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=>      0.00292  * 0.0499 =  C_2 * 1

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Generally from  Beer's law we have that

      A  =  \epsilon  * l  *  C_2

=>   \epsilon  =  \frac{A}{ l  *  C_2 }

Here  l is the length who value is  1 cm because the unit of  molar  absorptivity coefficient of the dye is M^{-1} *  cm^{-1}

So

            \epsilon  =  \frac{0.600}{ 1   * 0.0001457   }  

=>       \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

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Magnesium will give two electrons to oxygen and both will become ions.

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