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inna [77]
3 years ago
6

In which state of matter are water molecules measured as having a comparatively high temperature?

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

water vapor

because solid ice is a matter

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If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0
3 years ago
Active
klio [65]

Option B

Primary productivity is not limited by time

<u>Explanation:</u>

To ecosystems, the productivity of the primary producers is essential because certain bodies generate energy for different existence bodies. Besides the era, primary production ends in the extension of dissimilar plant biomass to the practice. Consumers acquire their strength from primary producers, unless directly (herbivores, any detritivores), or diffusely.

It depends superimposed on the availability of the sunlight, availability of the nutrients such as nitrogen, iron, phosphorus from the soil and water. The primary productivity of the plant's distinct autotrophs is not restricted by time.

5 0
3 years ago
8. If I have 34 grams of FeO, how many grams of Oxygen were synthesized?<br>Fe + O2 → Feo​
krek1111 [17]

Answer:

34 gram of FeO produced 8 gram of oxygen.

Explanation:

Given data:

Mass of FeO = 34 g

Mass of oxygen = ?

Solution;

Chemical equation:

2FeO → 2Fe + O₂

Number of moles of FeO:

Number of moles = mass/ molar mass

Number of moles = 34 g /71.8 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of FeO with oxygen:

             FeO       :       O₂

                2         :        1

                0.5      :      1/2 × 0.5 = 0.25

Mass of oxygen:

Mass = number of moles × molar mass

Mass =  0.25 mol × 32 g/mol

Mass = 8 g

So 34 gram of FeO produced 8 gram of oxygen.

8 0
3 years ago
Which regulates the flow of electrical current to the filament of the x-ray tube? a. high-voltage circuit
timurjin [86]
B) low-voltage circuit
8 0
2 years ago
One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc
Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

\text{The new partial pressure for }N_2 \ gas}

P_1V_1=P_2V_2

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar

\text{The new partial pressure for }Ar \ gas}

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar

Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = 2.225 \ Bar

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

2.1  \ bar = 2.07  \ atm \\ \\3.4 \  bar = 3.36 \  atm

For moles N₂:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}

n = 0.08297 \ mol  \ N_2

For moles of Ar:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}

n = 0.2172 \ mol  \ Ar

\mathtt{total \  moles = moles \ of \  N_2 + moles  \ of \ Ar}

=0.08297 mol + 0.2037 mol \\                   = 0.2867 mol gases

Finally;

The final pressure of the mixture is:

PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}

P = 2.217 atm

P ≅ 2.24 bar

7 0
3 years ago
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