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kap26 [50]
2 years ago
10

For the following systems (as underlined), determine which of the following conditions apply: open, closed, adiabatic, isolated,

isothermal, isobaric, isochoric, or steady-state. a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. (3 pts) b. The air inside the tire of a Nascar during the first minute of driving in a race. (3 pts) c. Your body over the last week.
Chemistry
1 answer:
zalisa [80]2 years ago
8 0

Answer:

a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. It is Closed because the freezer only exchanges energy, Isothermal since the freezer maintain the temperature constant and Isothermal and Isobaric because the ice cube remains with volume and pressure constant.

b. The air inside the tire of a Nascar during the first minute of driving in a race. Closed because the tire only exchange energy at first and Isochoric since the volume of the tire remain constant.

c. Your body over the last week. Open because the body exchange matter and energy.

Explanation:

The open, closed, adiabatic and isolated systems are defined considering if exchange matter or energy, as the definitions below:

- An <em>open</em> system exchange matter and energy.

- A <em>closed</em> system exchange only energy.

- An <em>adiabatic</em> system only exchange matter.

- An <em>isolated</em> system not exchange matter and energy

The isothermal, isobaric, isochoric, or steady-state are defined as follows:

- Isothermal is a process at a constant temperature.

- Isobaric is a process at constant pressure.

- Isochoric is a process at a constant volume.

- A steady-state refers to a reaction in which the concentrations of the reactants, intermediaries, and products don't change over time.

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Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0
2 years ago
Calculate Density:<br> Volume=13cm3, Mass=147.55g<br> Someone please help
almond37 [142]

Answer:

11.35 g/cm³

Explanation:

If your rounding then 11.4. hope this helps :)

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nadya68 [22]
Answer:
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Good luck !
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2 years ago
During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°
pantera1 [17]
Answer is: 39,083kJ.
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m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
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Q = 500g·18,7°C·4,18J/g·°C.
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3 years ago
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