One correct thging is that there are the same amount of positive and negative atoms
Answer:
¿Las partículas de hielo tienen una alta fuerza de atracción entre partículas? Justifica tu respuesta
Explanation:
Dichos contenidos están presentes en los currículos de Física y Química de la educación básica, con independencia del marco legal, pues introducen al alumno en el conocimiento químico de la materia. Aunque la teoría cinética molecular obvia la composición atómica de las partículas, no deja de ser un contenido deseable para introducir a los alumnos en el mundo de la química pues permite diferenciar y establecer relaciones entre los niveles macro, micro y simbólico de la materia.
The answer is: " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
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Note the exact conversion: 1000 mL = 1 L .
Given: 250 mL ;
250 mL = ? L ? ;
250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
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(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;
We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):
→ 0.50 mol CaCl₂ .
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1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
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There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
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So, to calculate the molar mass of "CaCl₂" :
40.08 g + 2(35.45 g) =
40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;
→ round to 111 g/mol .
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So:
→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;
→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;
= (0.50) * (111 g) CaCl₂ ;
= 55.5 g CaCl₂ ;
→ round to 2 significant figures;
→ 56 g CaCl₂ .
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The answer is: " 56 g CaCl₂ " .
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Molar volume is a property of a component in a solution. It is defined as the volume occupied by one mole of the component in the closed system. You would not expect all solutions to execute volume additivity because intermolecular forces between the components come into play. There is no such thing as conservation of volume.
Vapor pressure affects molar volume because gases are very sensitive by these process conditions. Vapor pressure is very temperature-dependent. Consequently, at a different temperature, your component could expand or compress, thus, affecting the molar volume. Moreover, the pressure affects the molecular collisions in the system.
Answer:
Mass = 18.9 g
Explanation:
Given data:
Mass of Al₂O₃ formed = ?
Mass of Al = 10.0 g
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 27 g/mol
Number of moles = 0.37 mol
Now we will compare the moles of Al and Al₂O₃.
Al : Al₂O₃
4 : 2
0.37 : 2/4×0.37 = 0.185 mol
Mass of Al₂O₃:
Mass = number of moles × molar mass
Mass = 0.185 mol × 101.9 g/mol
Mass = 18.9 g
