Answer:
Explanation:
During that process of being exposed to air and water while being left outside or in the elements for an extended period of time, a variety of different types of rusts can form, but the most common form is Fe2O3. Rust only forms on the outside of a metal surface because it requires exposure of oxygen and water to rust.
Existen 6 tipos de nutrientes: Glúcidos, Lípidos, Proteínas, Vitaminas, Agua y Sales minerales. Cada uno cumple unas funciones distintas, aportando los elementos necesarios para nuestras células.
Answer is: mass of the ore is 8.54kg.<span>
</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
