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Montano1993 [528]
2 years ago
14

Calculate the number of molecules in a deep breath of air whose volume is 2.15 L at body temperature, 37 ∘C, and a pressure of 7

40 torr .
Chemistry
2 answers:
Zinaida [17]2 years ago
7 0
PV = nRT 
R = 0.0821 L * atm / mol * K 
(ideal gas constant) 

First, convert 735 torr to atm. Divide by 760. 
(1 atm = 760 torr) 

735 torr * 1 atm / 760 torr = 0.967 atm 
Then, convert 37 C to Kelvin. Just add 273. 
37 C = 310K 

n = PV / RT 
= (0.967)(2.07) / (0.0821)(310) 
= 0.0786 mol 

<span>0.0786 mol * 6.02 * 10^23 molecules / 1 mol = 4.73 * 10^22 molecules </span>
ololo11 [35]2 years ago
6 0
PV=nRT
n= \frac{PV}{RT}
n= \frac{740 torr*2.15 L}{62.4 \frac{torr*L}{mol*K} *(37celsius+273)}
n = 0.0822 mol
0.0822 mol*6.022*10 ^{23}  = 4.95*10 ^{22} molecules
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Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

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K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
A sample of an ionic compound containing iron and chlorine is analyzed and found to have a molar mass of 126.8 g/mol. what is th
julsineya [31]
Answer is: <span>the charge of the iron in this compound is +2.
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Atomic mass of chlorine is 35,5 g/mol.
If compound is FeCl, molar mass would be 55,8 </span>g/mol + 35,5 g/mol = 91,3 g/mo, that is not correct.
If compound is FeCl₂, malar mass of compound would be:
55,8 g/mol + 2·35,5 g/mol = 126,8 g/mol, that is correct.
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5 0
3 years ago
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shusha [124]

Answer:

2.3 x 10-23 g.

Explanation:

The mass of a single atom is the mass number, 14, is the mass in grams of one mole of carbon.

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mass of 1 atom / 1 atom = mass of a mole of atoms / 6.022 x 10^23 atoms.

mass of 1 atom = mass of a mole of atoms / 6.022 x 1023

mass of 1 N atom = 14 / 6.022 x 10^23 N atoms

mass of 1 N atom = 2.325 x 10^-23 g

The mass of a single Nitrogen atom is 2.325 x 10-23 g.

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3 years ago
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The answer is b hope I helped
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(D) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants.

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