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1 month ago

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Suppose that a certain fortunate person has a net worth of $77.0 billion ($). If her stock has a good year and gains $3.20 billi

on () in value, what is her new net worth? Express your answer to three significant figures. Suppose that this individual now decides to give one-eighth of a percent (0.125 ) of her new net worth to charity. How many dollars are given to charity? The answer is not 10.0025 or 12.5. I have been working at this problem and just need to figure out how do this for future reference.

1 answer:

VMariaS [2.9K]1 month ago

3 0

Answer:

The revised net worth rounded to three significant figures is $80.2 billion.

The donation to charity amounts to $100,250,000.

Explanation:

Initial net worth is $77.0 billion.

Stock gains are $3.20 billion.

Calculating new net worth: $77.0 billion + $3.20 billion = $80.20 billion.

Hence, new net worth to three significant figures is $80.2 billion.

One-eighth of a percent is calculated as (1/8 x 1) / 100 = 0.00125.

The charitable contribution amounts to 0.00125 X $80.2 billion = $100,250,000.

Alternatively, $80,200,000,000 divided by 100 results in $802,000,000.

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The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft

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Answer: The air in the room weighs 37.068 kg

Explanation:

Given dimensions:

Length = 10.0 ft

Width = 11.0 ft

Height = 10.0 ft

The volume of the room (rectangular prism) is calculated using:

$V=lbh$

where l = length, b = breadth, h = height.

Substituting the values,

$V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L$

Using the conversion: $1ft^3=28.3168L$

Next, calculate the mass of the air based on density:

$Density=\frac{Mass}{Volume}$

$1.19g/L=\frac{Mass}{31148.53L}$

$Mass=37066.7507g=37.068kg$

Conversion: 1 kg = 1000 g

Thus, the air mass in the room equals 37.068 kilograms.

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2 months ago

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The electron configuration of an element is 1s22s22p4. Describe what most likely happens when two atoms of this element move tow

castortr0y [3046]

Answer:

Cuando dos átomos se acercan entre sí, se genera un compuesto al compartir pares de electrones que cada uno de los átomos aporta, permitiéndoles alcanzar los 8 electrones de valencia (octeto) en su capa externa.

Explanation:

La configuración electrónica del elemento puede escribirse de la siguiente manera;

1s²2s²2p⁴

La configuración electrónica dada es equivalente a la del oxígeno, por lo tanto, tenemos;

El número de electrones en la capa de valencia = 2 + 4 = 6 electrones

Por consiguiente, cada átomo necesita 2 electrones para completar sus 8 electrones (octeto) en la capa externa.

Al acercarse los dos átomos, reaccionan y se combinan para formar un compuesto al compartir 4 electrones, 2 de cada átomo, de modo que cada átomo obtenga 2 electrones adicionales en su órbita externa en el nuevo compuesto y así se logre la configuración estable de octeto para cada uno de los átomos en el compuesto recién formado.

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2 months ago

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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

eduard [2782]

Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

According to the De-Broglie equation $\lambda = \frac{h}{p}$

with p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$ = $3.6 \times 10^{-26} J/m$

Squaring both sides gives us:

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

$12.96 \times 10^{-52}$ = $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where m = mass of the electron

Therefore, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is $7.1 \times 10^{-22} J$ .

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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

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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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Answer:

The equation formulated by Michaelis-Menten is expressed as

v₀ = Kcat × [E₀] × [S] / (Km + [S])

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Explanation:

Refer to the attached image for an in-depth clarification

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1 month ago