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Olenka
27 days ago
10

What is the amount of heat released by 1.00 gram of liquid water at 0°C when it changes to 1.00 gram of ice at 0°C?

Chemistry
1 answer:
castortr0y [2.7K]27 days ago
8 0

Answer:

334 J/g

Explanation:

The relevant data provided in the question are as follows:

Mass (m) = 1 g

Specific heat of Fusion (Hf) = 334 J/g

Heat (Q) =?

By applying the formula Q = m·Hf, we can calculate the heat released:

Q = m·Hf

Q = 1 x 334

Q = 334 J

Thus, the total heat released amounts to 334 J.

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Three substances are poured in a cylinder with different densities. Draw the cylinder with the three layers and identify the sub
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Clarification:

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24 days ago
A sample of neon gas at STP is allowed to expand into an evacuated vessel. What is the sign of work for this process?
Alekssandra [2711]

Answer:

The work done in this process will be considered Negative.

Explanation:

The energy transferred by the system to the environment is negative

Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.

In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.

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3 0
1 month ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
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Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

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m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

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4 0
1 month ago
Read 2 more answers
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
KiRa [2711]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0
1 month ago
A student checks the air in her bicycle tires early in the morning when it is cool outside. If she measures it again later in th
Anarel [2600]

Answer:

She will likely notice an increase in tire pressure.

Explanation:

According to the ideal gas law, pressure is directly related to temperature. Therefore, as temperature rises, so does pressure:

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6 0
1 month ago
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