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2 years ago

A SCUBA diver with a 4-liter lung capacity is at a depth of 135 feet (4 atm.) She ascends to the surface where the

Chemistry

2 answers:

Nostrana [21]2 years ago

5 0

Answer:

16L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 4L

P1 (initial pressure) = 4atm

P2 (final pressure) = 1atm

V2 (final volume) =?

Using Boyle's law equation P1V1 = P2V2, the new volume can be obtain as follow:

P1V1 = P2V2

4 x 4 = 1 x V2

16 = V2

V2 = 16L

Therefore, her new lungs volume is 16L

anastassius [24]2 years ago

3 0

Answer:

the volume decreases, 16 Li hope i helped you good day and God bless!

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Mark the statement as true or false All of the proposed ideas by Dalton, Thompson, and Rutherford are still used in the modern t

kvv77 [185]

False

Although we use many of their ideas to describe atoms today, such as the existence of a tiny, dense nucleus in an atom (proposed by Rutherford), or the notion that all atoms of an element are identical (proposed by Dalton), some of their ideas have been rejected by the modern theory of the atom.

For example, Thompson came up with the plum pudding model to describe an atom, which resembled a sphere of positive charge with electrons embedded in it. We know now, however, that atoms are mostly empty space with a tiny, dense nucleus.

Another example is Dalton's atomic theory, which stated that atoms are indivisible particles. However, this was disproved by the discovery of subatomic particles.

4 0

2 years ago

Read 2 more answers

If you increase the solute amount but keep the solution volume the same what

solmaris [256]

Answer: Molarity increases

Explanation:

Molarity, also known as concentration in moles/dm3 or g/dm3, is calculated by dividing the amount of solute dissolved by the volume of solvent. So, Molarity (c) = amount of solute (n) / volume (v)

i.e c = n/v

Hence, molarity is directly proportional to the amount of solute dissolved, and inversely proportional to the volume of solvent.

Thus, at same volume, any increase in solute amount increases molarity while a decrease will also decreases molarity.

4 0

2 years ago

if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As

Black_prince [1.1K]

Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻

1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³

5 0

3 years ago

What is true about the element calcium(Ca, atomic number 20)? Choose all that apply.

Marat540 [252]

It has 2 valence electrons

THANK ME

7 0

3 years ago

Read 2 more answers

During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂

In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0

2 years ago