Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Answer:
vHe / vNe = 2.24
Explanation:
To obtain the velocity of an ideal gas you must use the formula:
v = √3RT / √M
Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:
vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol
vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
The ratio is:
vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol
<em>vHe / vNe = 2.24</em>
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I hope it helps!
<span>One mole of a substance contains Avogadro's number of atoms/molecules/the like. This would mean that all of the items described would have approximately 6.022 * 10^23 atoms, even though their masses would differ. This would be due to the molar mass of each substance being different because of the constituent elements in the substance.</span>
Answer:
A, T, C y G, son las "letras" del código del ADN; representan los compuestos químicos adenina (A), timina (T), citosina (C) y guanina (G), respectivamente, que constituyen las bases de nucleótidos del ADN. ...El código genético es el conjunto de reglas que define cómo se traduce una secuencia de nucleótidos en el ARNm a una secuencia de aminoácidos en una proteína
Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.
