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alexandr402 [8]
3 years ago
5

How many moles of air are there in a human lung with a volume of 2.4 L at STP? Explain your answer

Chemistry
1 answer:
ollegr [7]3 years ago
5 0
One mole of any gas at STP is about 22.4 liters (even for mixtures). This number comes from the ideal gas equation by setting n=1 and STP conditions:
V=nR/P =1 mole x 0.08206Latm/molK x 237K/1atm = 22.4L

so jsut divide: 893L/22.4L/mole = 39.9 moles
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The rate constant for a reaction is 4.65 L mol-1s-1. What is the overall order of the reaction? zero
il63 [147K]

Answer:

second order

Explanation:

units of reaction and their order.

Zero order --> M^1 s^-1 = M/s

First order --> M^0 s^-1 = 1/s

Second order --> M^-1 s^-1 = L/mol s

In the question rate constant k =  4.65 L mol-1s-1. = 4.65 L/mol s

Hence, the reaction is a second order reaction

8 0
3 years ago
Read 2 more answers
Which of the following is not a statement of Dalton's atomic theory of matter?a. Elements are made of atomsb. Atoms of a given e
Alona [7]

Answer:

c. The atoms of one element can be identical to the atoms of another element.

Explanation:

<em>Which of the following is not a statement of Dalton's atomic theory of matter?</em>

<em>a. Elements are made of atoms.</em> TRUE. An atom is the smallest particle of a chemical element that can exist.

<em>b. Atoms of a given element are identical.</em> TRUE. The only slight difference is in the mass of isotopes.

<em>c. The atoms of one element can be identical to the atoms of another element.</em> FALSE. The atoms of different elements are different from one to another.

<em>d. A given compound always has the same number and kinds of atoms. </em>TRUE. This is known as Dalton's law of constant composition.

3 0
3 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
3 years ago
Please help with this, i don't really get how to do this
LuckyWell [14K]
#4 and #5:
To find pH given concentration of H+ or H30+
pH = - log (H+ or H30+ M)

To find pH given concentration of OH-
Since you already found the pH for this (in #4), you subtract #4's answer from 14.
14 - (pH) = pOH
8 0
3 years ago
Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
aksik [14]
Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

<span>The anode is positive and the cathode is negative.</span>

7 0
3 years ago
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