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2 years ago

1. How much heat is required to melt 25.0 g of ice at 0°C?

Chemistry

1 answer:

tigry1 [53]2 years ago

8 0

Answer: The heat required to melt 25.0 g of ice at $0^0C$ is 8350 Joules

Explanation:

Heat of Fusion tells us how much energy is needed to convert 1g of a solid to a liquid at the same temperature.

$Q=m\times L$

Q = Heat absorbed = ?

m = mass of ice = 25.0 g

L = Latent heat of fusion of ice = 334 J/g

Putting in the values, we get:

$Q=25.0g\times 334J/g=8350J$

Thus heat required to melt 25.0 g of ice at $0^0C$ is 8350 Joules

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Sound waves are transverse mechanical waves. True False

allochka39001 [22]

False

Explanation:

Sound waves are longitudinal mechanical waves.

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A longitudinal wave is a wave that is directed parallel to their source.
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2 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

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1 year ago

Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H

murzikaleks [220]

Answer:

0.0184

Explanation:

Let's consider the following reaction at equilibrium.

2 HI(g) ⇌ H₂(g) + I₂(g)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂] × [I₂] / [HI]²

Kc = (4.78 × 10⁻⁴) × (4.78 × 10⁻⁴) / (3.52 × 10⁻³)²

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2 years ago

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Hello,

Thanks for posting your question here on brainly.

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2 years ago

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jarptica [38.1K]

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