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Mars2501 [29]
3 years ago
11

A bottle is partly filled with liquid water, sealed shut, and then placed under a heat lamp. The volume of the liquid water even

tually reaches a stable level because:
Chemistry
1 answer:
ANTONII [103]3 years ago
6 0

Because the rate at which water vapour condenses gets increase slowly to get equal to the rate of evaporation of the water.

Explanation:

When a bottle is partly filled with liquid water is leaves space for vapours to escape and get condensed equally.

When sealed and kept below the lamp the rate of condensation increases due to the empty space in the bottle for getting vapours cool down.

A point arrives when evaporation equates the condensation of the liquid in bottle becomes stable because vapours cannot pass the bottle eventually condense and become liquid.

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Difference between emprical and molecular formula
Musya8 [376]

Answer:

The key difference between empirical and molecular formulas is that an empirical formula only gives the simplest ratio of atom whereas a molecular formula gives the exact number of each atom in a molecule.

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The isotopes K-37 and K-42 have the same
Masteriza [31]

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the same number of protons

Explanation:

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3 years ago
Read 2 more answers
A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.
Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

  • Concentration of the weak acid (Ca): 0.187 M
  • pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}

3 0
2 years ago
What is the definition of pendulum
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weight hung from a fixed point so that it can swing freely backward and forward.

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100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
Olin [163]

Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

               = (M)(Volume of the solution (L))

               = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

7 0
3 years ago
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