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1 month ago

11

Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficient

s in the balanced equation. Do not forget coefficients of "one."
Cr2(SO4)3 + RbOH Cr(OH)3 + Rb2SO4

(a) 10
(b) 12
(c) 13
(d) 14
(e) 15

1 answer:

lorasvet [2.6K]1 month ago

6 0

Cr2(SO4)3 + 6RbOH --> 2Cr(OH)3 + 3Rb2SO4

1 + 6 + 2 + 3 equals 12 (b)

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When a heavy football player and a light one run into each other, which player hits the other with

Anarel [2643]

A heavier player collides with a lighter player using greater force.

The lighter player sustains more injuries following the impact.

Explanation:

A heavier player impacts a lighter player with greater intensity, resulting in more pronounced injuries to the lighter player post-collision.

Force is defined as mass multiplied by the acceleration of an object;

Force = mass x acceleration

We observe that as mass and acceleration increase, the force exerted rises accordingly.

Clearly, the heavier player's mass surpasses that of the lighter player, leading to a greater force exerted upon collision.

Moreover, the lighter player is likely to be injured more severely after the clash. The momentum generated by the heavier player during the impact is considerably significant. Once they collide, the lighter player will certainly alter their speed and trajectory.

Learn more:

Momentum

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1 month ago

A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h

lions [2702]

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution: 0.0328 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}$

The amount of HCl in the 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution: 0.0245 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}$

The amount of NaOH in the 0.100 L solution = 0.00245 moles

3) Determining the concentration of hydrochloric acid in the final solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl from the original 0.0082 moles of HCl.

The total volume of the mixture becomes 0.100 L + 0.250 L = 0.350 L

Remaining moles of unreacted HCl = 0.0082 moles - 0.00245 moles = 0.00575 moles

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$

Concentration of the remaining HCl: $\frac{0.00575 moles}{0.350L}=0.0164 M$

0.0164 molar concentration of hydrochloric acid in the resulting solution.

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26 days ago

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2459]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

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

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

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9 days ago

What fact do scientists believe provides evidence for the big bang theory?a. all comets follow elliptical orbits.b. most extra-s

alisha [2785]

Answer:

D

Reasoning:

This is due to the belief that at one time, the galaxy was condensed into a minute spot and the Big Bang event initiated the universe's expansion, propelling galaxies outward into space, distancing them from one another. Dark energy is theorized to exert the force from the Big Bang that drives this expansion.

8 0

18 days ago

Read 2 more answers

Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

lorasvet [2611]

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

$m_{tot}=m_{fuselage}+m_{passangers}$

Para el aluminio:

$m_{tot}=m_{fus-Al}+m_{pas-Al}$

$m_{fus-Al}=\delta _{Al}*V_{fuselage}$

y

$V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]$

donde:

L es longitud
D es diámetro
e es grosor

$m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}$

Para el acero (mismo procedimiento):

$m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel$

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

5 0

1 month ago