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2 months ago

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Sierra has a special kind of liquid rubber. She knows that ultraviolet light is absorbed by the rubber, X-ray light is transmitt

ed through the rubber, and blue light is reflected off the rubber. Sierra wonders if the liquid rubber will become solid if she shines the lights on it. Can light cause the rubber to become solid? Why or why not? Does it matter what type of light she shines on the rubber?

1 answer:

Anarel [2.9K]2 months ago

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You want to test the effect of pH on the distribution of Artemia in your 35cm long testing chamber. If you measure a pH = 1 at o

Tems11 [2777]

Answer:The pH measured 10 cm from the most acidic end is 3.42.

Explanation:

The pH at one end = 1The pH at the other end = 13

The chamber length = 13 cm

The change in pH concerning the chamber's length from the acidic end is

Thus, the pH at a distance of 10 cm from the most acidic end is 3.42.

$x=\frac{13-1}{35 cm}=\frac{12}{35} pH/cm$

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1 month ago

Which of the following statements reasonably explains why this reaction has a low activation energy? View Available Hint(s) The

eduard [2782]

Answer;

Considering the types of bonds being created and severed in the transition state, the stability of this temporary structure is comparatively high.

Explanation;

The reaction can be expressed as follows; NO(g)+F2(g)→NOF(g)+F(g)
All chemical reactions, including exothermic ones, require activation energy to initiate. The activation energy is the least amount of energy needed for the reactants to come together, overcome opposing forces, and begin breaking bonds.
When molecules encounter each other, their kinetic energy may be sufficient to stretch, bend, and eventually break bonds, resulting in chemical reactions.

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1 month ago

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

Tems11 [2777]

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.

3 0

1 month ago

What is the volume in mL of a 1.11 carat diamond, given the density of diamond is 3.51 g/mL? (1 carat = 200 mg). Use significant

eduard [2782]

To determine the length of each side,
employ the distance formula represented by the equation:
Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Calculating
<span>AB = 8 units   BC = 6 units AC = 10 units
</span><span>MN =8units    NO = 6 units MO = 10 units
</span><span>XY = 6.32 units   YZ = 6.32 units XZ = 8.94 units
</span>JK = 4.47 units    KL = 4.47 units JL = 6 units

1 The correct response is option b) triangles ABC and MNO are Congruent. <span>These triangles, ABC and MNO, have congruent side lengths.

</span>2 The answer is option c) rotation.
There is a rotation of 90º around the origin for triangles ABC and MNO, where B=N,
B=N
C----------O
A----------M

7 0

1 month ago

Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the

Tems11 [2777]

Answer:

THE MOLAR MASS OF XCL2 IS 400 g/mol

THE MOLAR MASS OF YCL2 IS 250 g/mol.

Explanation:

We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

Number of moles = mass / molar mass

Molar mass = mass / number of moles.

For XCl2,

mass = 100 g

number of moles = 0.25 mol

Thus, molar mass = mass / number of moles

Molar mass = 100 g / 0.25 mol

Molar mass = 400 g/mol.

For YCl2,

mass = 125 g

number of moles = 0.50 mol

Molar mass = 125 g / 0.50 mol

Molar mass = 250 g/mol.

Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.

3 0

2 months ago