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sammy [17]
3 years ago
7

Joebert had a rectangular cup and filled it with water. He measured the water's mass and volume. Then he got a density of 1.25g/

ml. If the accepted density of water is 1 g/ml, what was the percent error of his measurement?
someone can help me please?
Chemistry
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

Percent error = 25%

Explanation:

Given data:

Measured density of water = 1.25 g/mL

Accepted density value of water = 1 g/mL

Percent error = ?

Solution:

Formula:

Percent error = (measured value - accepted value / accepted value) × 100

Now we will put the values in formula:

Percent error = (1.25 g/mL - 1 g/mL /1 g/mL )× 100

Percent error = (0.25 g/mL /1 g/mL )× 100

Percent error = 0.25 × 100

Percent error = 25%

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An industrial vat contains 650 grams of solid lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with exc
mario62 [17]

Answer:

88.98 %.

Explanation:

  • From the balanced equation:

<em>2HCl + Pb(NO₃)₂ → 2HNO₃ + PbCl₂</em>

  • It is clear that 1.0 mole of Pb(NO₃)₂ reacts with 2.0 moles of HCl to produce 1.0 mole of PbCl₂ and 2.0 moles of HNO₃.
  • <em>The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100.</em>
  • The actual yield of lead(II) chloride (PbCl₂) = 650 g.
  • Now, we need to calculate the calculated yield of lead(II) chloride (PbCl₂).
  • We need to calculate the no. of moles (n) of lead(II) nitrate (Pb(NO₃)₂) (870 grams) using the relation: <em>n = mass / molar mass.</em>
  • n of lead(II) nitrate (Pb(NO₃)₂) = mass / molar mass = (870 g) / (331.2 g/mol) = 2.63 mol.
  • Since HCl is in excess, the limiting reactant is lead(II) nitrate (Pb(NO₃)₂).

<u><em>Using cross multiplication:</em></u>

1.0 mole of Pb(NO₃)₂ produces → 1.0 mole of PbCl₂, from the stichiometry.

∴ 2.63 mole of Pb(NO₃)₂ produces → 2.63 mole of PbCl₂.

  • The mass of PbCl₂ produced (the calculated yield) = n x molar mass = (2.63 mol) (278.1 g/mol) = 730.52 g.

∴ The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100 = [(650 g) / (730.52)] x 100 = 88.98 %.

5 0
3 years ago
Read 2 more answers
Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
mr_godi [17]

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
4 0
3 years ago
Can someone answer this question? It's not that long
Kitty [74]

Answer:

sections 2 and 3

Explanation:

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6 0
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If 1.0 gram of hydrogen reacts with 19.0 grams of fluorine, then what is the percent by mass of fluorine in the compound that is
astraxan [27]
The reaction between hydrogen (H2) and fluorine (F2) is given below,
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One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%. 
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3 years ago
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In terms of entropy and energy, systems in nature tend to undergo changes toward
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