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Roman55 [17]
3 years ago
15

The smallest part of a element that retains the properties of that element

Chemistry
1 answer:
rusak2 [61]3 years ago
5 0
An atom, it is obviously the smallest
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true because the bonds cannot be broken down

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What would be the effect on the observed melting point of sample were poorly packed?
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If a sample is packed poorly, the sample will not heat evenly and will take longer to melt.
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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne
Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0
3 years ago
the pressure in a sealed plastic container is 108 kPa at 41 degrees Celsius. What is the pressure when the temperature drops to
Katarina [22]

<u>Answer:</u> The new pressure will be 101.46 kPa.

<u>Explanation:</u>

To calculate the new pressure, we use the equation given by Gay-Lussac Law. This law states that pressure is directly proportional to the temperature of the gas at constant volume.

The equation given by this law is:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are initial pressure and temperature.

P_2\text{ and }T_2 are final pressure and temperature.

We are given:

By using conversion factor:   T(K)=T(^oC)+273

P_1=108kPa\\T_1=41^oC=314K\\P_2=?kPa\\T_2=22^oC=295K

Putting values in above equation, we get:

\frac{108kPa}{314K}=\frac{P_2}{295K}\\\\P_2=101.46kPa

Hence, the new pressure will be 101.46 kPa.

3 0
3 years ago
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