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3 years ago

6

According to the graph, what is the name for the change in water temperature that occurs at approximately 1300 m? A. halocline B

. thermocline C. isocline D. baryocline

1 answer:

iVinArrow [24]3 years ago

4 0

The answer is b thermocline

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10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?

il63 [147K]

Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!

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3 years ago

Scientific explanations

Sergio039 [100]

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3 years ago

How many moles of methane are in 20.32 x 10^16 molecules

Marizza181 [45]

Taking into account the definition of avogadro's number, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

First of all, you have to know that Avogadro's number indicates the number of particles of a substance (usually atoms or molecules) that are in a mole.

Its value is 6.023×10²³ particles per mole and it applies to any substance.

Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of methane, then 20.32×10¹⁶ molecules are contained in how many moles of methane?

amount of moles of methane= (20.32×10¹⁶ molecules × 1 mole)÷ 6.023×10²³ atoms

Solving:

<u><em>amount of moles of methane= 3.37×10⁻⁷ moles</em></u>

Finally, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

Learn more about Avogadro's Number:

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2 years ago

The model below represents a nuclear reaction. Atom 1 is present before the reaction and Atom 2 and Atom 3 are produced after th

belka [17]

The right answer for the question that is being asked and shown above is that: "d. does not produce energy in nuclear power plants." The model most likely represents a reaction which d. does not produce energy in nuclear power plants<span>
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2 years ago

Read 2 more answers

The colligative molality of an unknown aqueous solution is 1.56 m.

yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- <u>Vapor pressure lowering</u>

ΔP = P° . Xm . i

- <u>Boiling point elevation</u>

ΔT = Kb . m . i

-<u> Freezing point depressión</u>

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

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3 years ago