TNT has the molecular formula: C7H5N3O6. And hence, when reacted in oxygen gas, you get what is known as <span>combustion</span> reaction. the reaction is: <span><span>C7</span><span>H5</span><span>N3</span><span>O6</span>+<span>O2</span>→C<span>O2</span>+<span>N2</span>+<span>H2</span><span>O</span></span>
What are the following examples?
Respuesta:
199.5 g
Explicación:
Paso 1: Escribir la reacción balanceada
2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂
Paso 2: Calcular la masa pura de 50 g de Al
Aluminio tiene 10% de impurezas, es decir, 10% de 50 g = 5 g. Luego, tiene 50 g - 5 g = 45 g de Al puro.
Paso 3: Calcular la masa teórica de Al₂(SO₄)₃ obtenida a partir de 45 g de Al
La relación de masas de Al₂(SO₄)₃ a Al es 342:54.
45 g Al × 342 g Al₂(SO₄)₃/54 g Al = 285 g Al₂(SO₄)₃
Paso 4: Calcular la masa real de Al₂(SO₄)₃ obtenida
El rendimiento de la reacción es de 70%.
285 g × 70% = 199.5 g
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
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3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>
Answer:
2
Explanation:
It is going through the air and is moving so that means that it has kinetic energy. But it is also going to go the down, which gives it potential energy.
