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elixir [45]
3 years ago
10

Which of the following fractions can be used in the conversion of 1500 dm3 to cm3? (2 points) fraction cubed: ten to the negativ

e first power m over one dm fraction cubed: ten to the negative second power cm over one dm fraction cubed: one cm over ten to the negative second dm fraction cubed: one meter over ten to the negative first power dm
Chemistry
1 answer:
Lina20 [59]3 years ago
6 0
We need to identify the fractions that can be used in converting dm³ to cm³ but first, we need to know what is dm and cm. The dm unit is decimeter and cm unit is for centimeter. We also need to take note that 1 decimeter is equal to 10 centimeters. Thus, the conversion of cubed dm to cubed cm is shown below:
1500 dm³ * (10 cm / 1 dm)³ , we distribute cube
= 1500 dm³ * 10³cm³ / dm³
= 1500 dm³ * 1000 cm³ / dm³, we cancel dm³ unit and the only cm³ will be left out such as:
= 1,500,000 cm³ 

The answer is the last in the choices which is "1 meter over 10 to the first power dm".
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Answer:

THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g

Explanation:

In an ideal condition

PV = nRT or PV = MRT/ MM where:

M = mass = unknown

MM =molar mass = 28 g/mol

P = pressure = 2 atm

V = volume = 25 mL = 0.025 L

R = gas constant = 0.082 L atm/mol K

T = temperature = 290 K

n = number of moles

The gas in the question is nitrogen gas

Molar mass of nitrogen gas = 14 * 2 = 28 g/mol

Then equating the variables and solving for M, we have

M = PV MM/ RT

M = 2 * 0.025 * 28 / 0.082 * 290

M = 1.4 / 23.78

M = 0.0589 g

The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

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\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

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