M = n/V
.5M = n/.100 L
n = .1 L * .5M
n= .05 mols of MgCl2
mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2
mass of MgCl2 = 4.76 grams
4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
Liquid silver is less dense than solid silver, so the solid silver would sink
The HNO3 is considered to be a Bronsted - Lowry acid, when this substance 'HNO3', will donate a proton, then it will form another substance. It will form two substances when the proton is donated in the water molecule. The two substances that will be formed is a nitrate iron and a hydronium ion.
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
-245.7°C es la temperatura del gas bajo 100 torr
Explanation:
Para resolver esta pregunta debemos hacer uso de la ley de Boyle que establece que la presión de un gas es directamente proporcional a la temperatura de este cuando el volumen permanece constante. La ecuación es:
P1T2 = P2T1
<em>Donde P es presión y T temperatura absoluta del estado inicial, 1, y final, 2.</em>
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Reemplazando:
P1 = 1000torr
T2 = ? -Incógnita-
P2 = 100torr
T1 = 273K -Temperatura del hielo fundido = 0°C = 273K
1000torrT2 = 100torr273
T2 = 27.3K
27.3K - 273 =
<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>
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