A 0.500-mole sample of a gas has a volume of 11.2 liters at 273 k. what is the pressure of the gas? (hint: use ideal gas law equ

Peter performed an experiment to test which of the three compounds his teacher gave him was an ionic compound. The table shows h

lions [2927]

B is identified as an ionic compound. Explanation: This conclusion is drawn based on the characteristics of ionic compounds, which typically form crystals similar to salt (sodium chloride), are soluble in water, exhibit high boiling and melting points, possess hardness from strong intermolecular forces, and can conduct electricity—though only when dissolved or in a molten state.

5 0

1 month ago

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You have a balloon filled with hydrogen gas which keeps it at a

lorasvet [2795]

The resulting temperature is 46.5°C.

Details:

According to Charles's law, the volume of gas, while maintaining constant pressure, correlates directly with temperature in Kelvin.

The formula representing Charles's law is expressed as follows:

$$\frac{V}{T} = constant$

$$\frac{V1}{T1} = \frac{V2}{T2}$

We need to determine T2, thus:

$$T2 = \frac{V2T1}{V1}$

V1 = 736 ml = 0.736 L

T1 = 15 ° C

V2 = 2.28 L

Substituting the values gives us:

T2 = $$\frac{2.28 \times 15}{0.736}$

= 46.5°C

It is evident that as the volume increases, the temperature also rises.

5 0

2 months ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

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2 months ago

The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil

lorasvet [2795]

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Provided information

Average separation between oxygen and nitrogen atoms: 115 pm

Step 2: Change the distance to meters (SI standard unit)

Using the conversion 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Transform the distance to millimeters

Employing the conversion 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

5 0

1 month ago

Which of the following statements is true about the relationships between photon energy, wavelength, and frequency?

KiRa [2933]

Answer: The Answer is A.

Explanation:

The energy of a photon is directly related to its electromagnetic frequency, meaning it is inversely related to the wavelength. A higher frequency results in greater energy for the photon. Conversely, a longer wavelength corresponds to lower energy levels.

Hope this Helps!

8 0

2 months ago