Answer:
24.5
Explanation:
15/6.3 => 2.38 half lives passed.
.5^2.38 => 0.19198 decimal representation of the percentage that is left over after 2.38 half lives have passed.
0.19198 *128 = 24.5 mg of the material remaining.
Answer:
-1190.24 kJ
Explanation:
The enthalpy change in a chemical reaction that produces or consumes gases is given by the expression:
ΔH = ΔU + Δngas RT
where Δn gas is the change of moles of gas, R is the gas constant,and T is temperature.
Now from the given balanced chemical reaction, the change in number of mol gas is equal to:
Δn gas = mole gas products - mole gas reactants = 2 - 5/2 = -1/2 mol
Sionce we know ΔU and the temperature (298 K), we are in position to calculate the change in enthalpy.
ΔH = -1189 x 10³ J + (-0.5 mol ) 8.314 J/Kmol x 298 K
ΔH = -1.190 x 10⁶ J = -1.190 x 10⁶ J x 1 kJ/1000 J = -1.190 x 10³ J
Answer:
first let's find the molar mass of CH2.Molar mass CH2=12.0+(2multiply1.01)=14.02g/mol
Explanation:
Respuesta:
21.8 L
Explicación:
Paso 1: Escribir la reacción balanceada
CaCO₃ ⇒ CaO + CO₂
Paso 2: Convertir 450 g de CaCO₃ a moles
La masa molar de CaCO₃ es 100.09 g.
450 g × (1 mol/100.09 g) = 4.50 mol
Paso 3: Calcular los moles de CO₂ que se forman a partir de 4.50 moles de CaCO₃
La relación molar de CaCO₃ a CO₂ es 1:1. Los moles de CO₂ formados son 1/1 × 4.50 mol = 4.50 mol.
Paso 4: Convertir la temperatura a Kelvin
Usaremos la siguiente expresión.
K = °C + 273.15 = 200°C + 273.15 = 473 K
Paso 5: Calcular el volumen de CO₂
Usaremos la ecuación del gas ideal.
P × V = n × R × T
V = n × R × T / P
V = 4.50 mol × (0.082 atm.L/mol.K) × 473 K / 8 atm
V = 21.8 L
<span>You are given O2 and C3H8, this is a combustion
reaction. The chemical reaction is C3H8 + 10O2 à 3CO2 + 4H2O. You are also given the molar mass
of O2 which is 32.00 g/mol and C3H8 which is 44.1 g/mol. You are required to
find the mass of O2 in grams. Since you have the reaction, oe mole of C3H8 is
required to completely react 10 moles of O2. So,</span>
0.025g C3H8(1 mol C3H8/44.1 g C3H8)(10 mol O2/1
mol C3H8)(32 g O2/1 mol O2) = <u>0.1802 g O2
</u>
<span> </span>
