Complete question:
What mass of steam initially at 120 ⁰C is needed to warm 200g of water in a 100 g glass container from 20.0 oC to 50.0 ⁰C
Answer:
the initial mass of the steam is 10.82 g
Explanation:
Given;
mass of water, m₁ = 200 g
mass of the glass, m₂ = 100 g
temperature of the steam = 120 ⁰C
initial temperature of the water, 20⁰ C
final temperature of the water, = 50⁰ C
let the mass of the steam = m
specific heat capacity of water c = 1 cal/g ⁰ C
specific heat capacity of glass c₂ = 0.2 cal/g ⁰ C
laten heat of vaporization of steam L = 540 cal/g
Apply principle of conservation energy;
Heat given off by the steam = Heat absorbed by water + heat absorbed by glass
[tex]mc\Delta T_1 + mL + mc\Delta T_2 = m_1c\Delta T_3 + m_2c_2\Delta T_3\\\\mc\Delta T_1 + mL + mc\Delta T_2 = [m_1c + m_2c_2]\Delta T_3[/tex]
m(1) (120 - 100) + m(540) + m(1) (100 - 50) = [200(1) + 100(0.2)] (50 - 20)
20m + 540m + 50m = 6600
610 m = 6600
m = 6600 / 610
m = 10.82 g
Therefore, the initial mass of the steam is 10.82 g
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
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
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%Que es la actividad física y en qué mejora
