Answer: the molecular formula of trioxide is ClOClO3 or Cl2O4
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A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1
Answer:
"1.4 mL" is the appropriate solution.
Explanation:
According to the question,
[tex]v_0=500[/tex][tex]\alpha =1.12\times 10^{-4}[/tex][tex]\Delta \epsilon = 25[/tex]Now,
Increase in volume will be:
⇒ [tex]\Delta V = \alpha\times v_0\times \Delta \epsilon[/tex]
By putting the given values, we get
[tex]=1.12\times 10^{-4}\times 500\times 25[/tex]
[tex]=1.12\times 10^{-4}\times 12500[/tex]
[tex]=1.4 \ mL[/tex]
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
Que es la actividad física y en qué mejora
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%A major component of gasoline is octane when octane is burned in air it chemically reacts with oxygen to produce carbon dioxide and water what mass of carbon dioxide is produced by the reaction of oxygen
gasoline is the chemical that is coming out of the air
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
