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2 months ago

According to reference table adv-10, which reaction will take place spontaneously?

1 answer:

5 0

Missing table!! list out the elements with their symbols starting with capital letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The oxidation potential of Pb(s) --> Pb2+(aq) is 0.126 V
The reduction potential for Ni2+(aq) --> Ni(s) is -0.25 V

Summing these gives a potential of -0.124 V (NOT SPONTANEOUS, AS THE SIGN IS NEGATIVE)

Au3+ + Al(s) → Au(s) + Al3+ Au3+(aq) --> Au(s) +1.5 V Al --> Al3+ +1.66 VV= 3.16 (SPONTANEOUS, POSITIVE SIGN)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn --> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu --> Cu2+ V = -0.337V

V= -0.777V (no spontaneous)

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The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

eduard [2782]

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

8 0

2 months ago

Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d

KiRa [2933]

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

$K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2$ (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

$v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}$

$k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}$

The spring constant calculates to 23045 N/m

3 0

2 months ago

In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a

KiRa [2933]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

$\Delta H =\frac{Q}{n}$

To calculate the heat (Q):

$Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J$

Next, we need to calculate the number of moles involved in the reaction (n):

$n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles$

With these two values, we can substitute them into the first equation:

$\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole$

Have a great day!

5 0

2 months ago

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

eduard [2782]

Answer:

The volume of calcium hydroxide solution utilized is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

In accordance with the reaction, 2 moles of KHP react with 1 mole of calcium hydroxide, thus 0.0330 moles of KHP will react with;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

The molarity of calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

The volume of the calcium hydroxide solution utilized is 0.0235 mL.

4 0

2 months ago

A girl stands on a bathroom scale. In metric units, the scale indicates that her weight is 42

VMariaS [2998]

Explanation:

The scale under her feet exerts an equal force but opposite in direction.

This principle aligns with Newton's third law of motion, which says that "for every action, there is an equal and opposite reaction."

As the girl’s weight presses down on the scale with a force of 42N, the scale responds with an equal upward force in the opposite direction.

The resulting net force is zero, which explains why her weight does not break the scale.

Learn more:

Force

7 0

2 months ago