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kiruha
1 month ago
13

A 100 milliliters (mL) sample of a liquid is poured from a beaker into a graduated cylinder. Which of the following best explain

s what happened to the liquid?
Chemistry
1 answer:
castortr0y [3K]1 month ago
7 0

Answer:

This question lacks completeness

Explanation:

This inquiry is indeed incomplete. However, when a specific volume of liquid is transferred from one container (like a beaker) to another (such as a graduated cylinder), the apparent volume decreases slightly. This occurs because some water molecules may adhere to the inner surface of the transferring container (the beaker). Thus, the amount of liquid will be marginally less than 100 mL in the graduated cylinder.

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Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha
VMariaS [2998]
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

The empirical formula is C₂H₅.
7 0
1 month ago
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
Alekssandra [3086]

Respuesta:

El oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

La oxidación se define como la pérdida de electrones. La oxidación provoca un aumento en el número de oxidación de un elemento.

Si se descompone esta reacción en sus mitades de reducción y oxidación

Se observa que, de los reactivos mencionados anteriormente,

H202 se convierte en H2O y O2

MnO4- + H+ se convierte en Mn2+ y H2O

El número de oxidación de Mn cambia de +7 en MnO4- a +2 en Mn2+ (lo que indica evidentemente una reducción)

El oxígeno en MnO4- no cambia su número de oxidación, ya que se mantiene en -2

El número de oxidación del oxígeno cambia de -1 en H2O2 a -2 en H2O y 0 en O2

El hidrógeno en H2O2 no cambia su número de oxidación, y su número de oxidación se mantiene en +1 tanto en H2O2 como en H2O.

Esto indica que H2O2 sufre tanto oxidación como reducción; más específicamente, el oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Espero que esto ayude

7 0
1 month ago
Why is it important to have regular supervision of the weight and measurements in the market
lions [2927]

Answer:

Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.

Explanation:

4 0
2 months ago
PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat
alisha [2963]

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

DHAP\rightleftharpoons G_3P

  • For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

\Delta G^o=-RT\ln K_{eq}

The free energy change can be calculated using the following equation:

\Delta G=\Delta G^o+RT\ln Q

Or,

\Delta G=-RT^o\ln K_{eq}+RT\ln Q

where,

\Delta G = Change in free energy

R = Gas constant = 8.314J/K mol

T^o = standard temperature = 25^oC=[273+25]K=298K

T = temperature of the cell = 37^oC=[273+37]K=310K

K_[eq} = equilibrium constant = 5.4\times 10^{-2}

Q = reaction quotient = \frac{[G_3P]}{[DHAP]}

[G_3P] = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol

Thus, the change in free energy for the reaction is -5339.76 J/mol

  • For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, -5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol

Consequently, the free energy change's units are kJ/mol

  • For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

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1 month ago
The chemical formula for cesium oxide is Cs2O. What is the charge of cesium? +1 +2 –1 –2
lions [2927]
The charge on cesium is +1
5 0
2 months ago
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