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Karo-lina-s [1.5K]
3 years ago
5

Given the following skeleton equation, what are the coefficients for the chemicals in the reaction? __H20 + __ P4O10 = __ H3PO4

Chemistry
1 answer:
AysviL [449]3 years ago
4 0
6H2O + P4O10 = 4H3PO4
Coefficients: 6, 1, 4
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he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer
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Answer:

\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

\Delta S=\frac{n*\Delta H}{T}

Whereas n accounts for the moles which are computed below:

n=5.9g*\frac{1mol}{72g} =0.082mol

Thus, the entropy turns out:

\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Best regards.

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3 years ago
What is a change in a substance that does not involve the identity of the substance
GenaCL600 [577]

The answer to this would be a physical change. Physical changes are changes that affect the form of a chemical substance, but not the chemical composition itself. Hope this helped!


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If 97.3 L of NO2 forms measured at 35 C and 632 mm Hg. What is the percent yield?
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<span>a. Use PV = nRT and solve for n = number of mols O2. 
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 Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR. 

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</span>and % will be 60%.
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3 years ago
Calcule el volumen de disolución de LiOH a 3.5 M, necesario para neutralizar una disolución de 25 ml de H2SO3 cuya densidad es d
Olin [163]

Respuesta:

0.11 L

Explicación:

Paso 1: Escribir la ecuación balanceada

2 LiOH + H₂SO₃ ⇒ Li₂SO₃ + H₂O

Paso 2: Calcular la masa de solución de H₂SO₃

25 mL de solución de H₂SO₃ tiene una denisdad de 1.03 g/mL.

25 mL × 1.03 g/mL = 28 g

Paso 3: Calcular la masa de H₂SO₃ en 26 g de Solución de H₂SO₃

La riqueza de H₂SO₃ es 60%, es decir, cada 100 g de solución hay 60 g de H₂SO₃.

26 g Sol × 60 g H₂SO₃/100 g Sol = 16 g H₂SO₃

Paso 4: Calcular los moles correspondientes a 16 g de H₂SO₃

La masa molar de H₂SO₃ es 82.07 g/mol.

16 g × 1 mol/82.07 g = 0.19 mol

Paso 5: Calcular los moles de LiOH que reaccionan con 0.19 moles de H₂SO₃

La relación molar de LiOH a H₂SO₃ es 2:1. Los moles de LiOH que reaccionan son 2/1 × 0.19 mol = 0.38 mol.

Paso 6: Calcular el volumen de solución de LiOH

0.38 moles de LiOH están en una solución 3.5 M. El volumen requerido es:

0.38 mol × 1 L/3.5 mol = 0.11 L

7 0
3 years ago
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