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Lelu [443]
2 years ago
6

Variation definition

Chemistry
1 answer:
mote1985 [20]2 years ago
4 0

Answer:

Explanation:

Variation problems involve fairly simple relationships or formulas, involving one variable being equal to one term. Here follows the most common kinds of variation.

The constant of variation in a direct variation is the constant (unchanged) ratio of two variable quantities. In the following equation y varies directly with x, and k is called the constant of variation:

y=kx

Another form of variation is the inverse variation which works when there is a relationship between two variables in which the product is a constant. When one variable increases the other decreases in proportion so that the product is unchanged.

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What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O
Ipatiy [6.2K]

Answer:

Mass = 13.23 g  

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂     →      3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

              O₂           :          C₃H₈

               5            :            1

             1.5            :          1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g  

6 0
2 years ago
A sample of nitrogen gas had a volume of 500. ml, a pressure in its closed container of 740 torr and a temperature 25 degrees c
Semenov [28]
Question:
              <span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?

Answer:

Data Given:
                   V</span>₁  =  500 mL

                   P₁  =  740 torr

                   T₁  =  25 °C + 273  =  298 K

                   V₂  =  ?

                   P₂  =  760 torr

                   T₂  =  50 °C + 273  =  323 K

Solution:
             Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,

                      P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                      V₂  =  (P₁ V₁ T₂) ÷ (T₁ P₂)

Putting Values,
  
                      V₂  =  (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)

                      V₂  =  527.68 mL
7 0
3 years ago
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