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2 years ago

Lovemyselftruly

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Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo s

Ray Of Light [21]

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como: P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:

$\frac{V}{T}=k$

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

$\frac{P}{T}=k$

Combinado las mencionadas tres leyes se obtiene:

$\frac{P*V}{T} =k$

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

P1: 750 torr
V1: 8.5 L
T1: 20°C= 293°K (siendo 0°C=273°K)
P2: 425 torr
V2: ?
T2: -20°C= 253 °K

Reemplazando:

$\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}$

Resolviendo:

$V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}$

V2= 12.95 L

El volumen del gas era 12.95 L

5 0

2 years ago

An experiment requires that enough C3H8O be used to yield of oxygen .

blsea [12.9K]

Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).

Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.

8 0

3 years ago

Once an enzymatic reaction is completed, the enzyme releases what?

sleet_krkn [62]

Answer:

Once an enzymatic reaction is completed, the enzyme releases substrates.

Explanation:

The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).

5 0

2 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

A student uses a balance to determine the mass of an object as 5.719 grams. the correct mass of the object is 5.586 grams. what

zaharov [31]

The percentage error is given by multiplying relative error by 100%.
Therefore, to get the percentage error we need relative error which is given by dividing the absolute error with the actual value.
Absolute error = 0.133
Percentage error = 0.133/5.586 × 100%
= 2.38%
The percentage error is therefore; 2.38%

8 0

2 years ago