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3 years ago

How many moles of oxygen will occupy a volume of 2.50 liters?

1 answer:

7 0

Data:
n (number of mols) = ?
V (volume) = 2.50 Liters
If:
1 L → 1000 g
2.50 L → y

y = 1000*2.50 = 2500 g

Therefore:
m (mass) = 2500 g

Now:
Molar Mass (MM) of oxygen = 16 g/mol

Formula:
$n = \frac{m}{MM}$

Solving:
$n = \frac{m}{MM}$
$n = \frac{2500\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol}$
$\boxed{\boxed{n = 156.25\:mols}}\end{array}}\qquad\quad\checkmark$

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The following are the main steps in the generation of the action potential.1. Sodium channels are inactivated2. Voltage gated po

Eduardwww [97]

Answer:

E) 4,6,7,1,2,3,5

Explanation:

The correct sequence of events is as follows.
A grated depolarization brings an area of excitable membrane to a threshold.
Sodium channels are activated and open, allowing positively-charged sodium ions to pass through.
The sodium ions then enter the cell, and depolarization of the negatively-charged cell membrane is initiated.
Sodium channels are first inactivated.
The voltage-gated potassium channels then open, allowing for potassium ions to move out of the cell, initiating repolarization.
Sodium channels then regain their normal properties allowing the sodium ions to pass through and making the cell membrane repolarized again.
The cell membrane will reach its resting membrane potential again.
Afterward, a temporary hyperpolarization occurs.

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3 years ago

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Design a concept map that shows the relationships among the following terms: volume , derived unit , mass , base unit , time , a

Svetlanka [38]

Base unit is the fundamental unit of measurement that is not mutually dependent. The seven base units are,

Length - meter

Mass- kilogram

Time - second

Temperature - Kelvin

Current - Ampere

Luminous intensity - Candela

Amount of matter - mole

Derived units are derived from the seven base units. Like the area (square meters) is the square of length (meters), volume is the cube of length.

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3 years ago

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

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3 years ago

In an experiment, dilute hydrochloric acid reacts with a white solid and bubbles of gas are produced. If this gas is bubbled thr

Andrej [43]

Answer:

Explanation:

Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white.

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3 years ago

What is the final pH of a solution obtained by mixing 300 ml of 0.4 M NH3 with 175 ml of 0.3 M HCl? (Kb = 1.8 x 10-5) Show all o

MariettaO [177]

Answer:

pH of the final solution = 9.15

Explanation:

Equation of the reaction: HCl + NH₃ ----> NH₄Cl

Number of moles of NH₃ = molarity * volume (L)

= 0.4 M * (300/1000) * 1 L = 0.12 moles

Number of moles of HCl = molarity * volume (L)

= 0.3 M * (175/1000) * 1 L = 0.0525 moles

Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced

Number moles of NH₄Cl produced = 0.0525 moles

Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675

pOH = pKb + log([salt]/[base])

pKb = -logKb

pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)

pOh = 4.744 + 0.109

pOH = 4.853

pH = 14 - pOH

pH = 14 - 4.853

pH = 9.15

Therefore, pH of the final solution = 9.15

3 0

3 years ago