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3 years ago

about the lung with balloons what respiratory organ do the small balloons represents? the big balloons?

Chemistry

1 answer:

butalik [34]3 years ago

7 0

Answer:

The plastic bottle represents the chest cavity, the straws act as the bronchi, the small balloons inside the plastic bottle serves as the lungs and the big balloon at the bottom of the plastic bottle is the diaphragm.

Explanation:

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Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$

[HB] is the concentration of base.

$8 * 10^{-5} =\frac{x^{2} }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1$

$x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}$

Then

$pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54$

So the pH of the solution is 2.54.

4 0

3 years ago

Electrons involved in bonding between atoms are____.

mariarad [96]

Answer:

valence electrons

8 0

3 years ago

A 3.93 g sample of a laboratory solution contains 1.25 g of acid. what is the concentration of the solution as a mass percentage

WITCHER [35]

Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%

4 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

2 years ago

You have to prepare 100.0 mL of a 0.100 M solution of sodium carbonate. You have a concentrated solution of sodium carbonate tha

Georgia [21]

Answer:

6.9 ml of concentrate

Explanation:

100 ml of .1 M will require .01 moles

from a 1.45 M solution, .01 mole would be

.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml

4 0

1 year ago