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2 days ago

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Exactly 3.705kg of substance Y are needed to neutralise 100 moles of HCL aq. What could be substance Y ? A. Mgcl2 and Caco3 B. M

gco3 and Ca(NO3)2 C. Mg(NO3)2 and Ca(NO3)2 D. Mgo and Cao

1 answer:

eduard [944]2 days ago

5 0

Correct answer:

C) Ca(OH)2

Note: The listed options in the question are incorrect. The accurate selections are shown below:

15. A total of 3.705 kg of substance Y is required to neutralize 100 moles of HCl(aq).

What might substance Y be? A Ca B CaO C Ca(OH)2 D CaCO3

Explanation:

A) Ca + 2HCl ---> CaCl2 + H2

Molar mass of Ca = 40 g/mol

The mass of Ca needed to neutralize 100 moles of HCl = 1/2 * 100 * 40 g = 2000g or 2 kg

B) CaO + 2HCl ---> CaCl2 + H2O

Molar mass of CaO = 56.1 g/mol

The mass of CaO needed to neutralize 100 moles of HCl = 1/2 * 100 * 56.1 g = 2805 g or 2.805 kg

C) Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

Molar mass of Ca(OH)2 = 74.1 g/mol

The mass of Ca(OH)2 necessary to neutralize 100 moles of HCl = 1/2 * 100 * 74.1 g = 3705 g or 3.705 kg

D) CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Molar mass of CaCO3 = 100 g/mol

The mass of CaCO3 required to neutralize 100 moles of HCl = 1/2 * 100 * 100 = 5000 g or 5.00 kg

Thus, the right choice is C.

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Fermium-253 is a radioactive isotope of fermium that has a half-life of 3.0 days. A scientist obtained a sample that contained 2

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Problem 2

You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.

Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.

Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.

#days Amount in micrograms

0 216

3 108

6 54

9 27

Problem One

Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.

Table

Bond Energy Kj/Mol Bond Length pico meters

N - N 167 145

N=N 418 125

N≡N 942 110

Rules

As the number of bonds INCREASES, the energy within the bond also INCREASES

As the number of bonds INCREASES, the distance of the bond DECREASES.

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8 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 [960]

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.

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14 days ago

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a

castortr0y [927]

Answer:

A) The molecular formula for ibuprofen is $C_{13}H_{18}O_2$

B) The molecular formula for Cadaverine is $C_{5}H_{14}N_2$

C) The molecular formula for Epinephrine is $C_9H_{13}O_3N_1$

Explanation:

Element percentage in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is = $C_xH_yO_z$

Count of carbon atoms in one ibuprofen molecule;

$75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100$

$x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13$

Count of hydrogen atoms in one ibuprofen molecule;

$8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100$

$y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18$

Count of oxygen atoms in one ibuprofen molecule;

$15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100$

$z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2$

Molecular formula for ibuprofen:

= $C_xH_yO_z= C_{13}H_{18}O_2$

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is = $C_xH_yN_z$

Count of carbon atoms in one Cadaverine molecule;

$58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100$

$x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5$

Count of hydrogen atoms in one Cadaverine molecule;

$13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100$

$y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14$

Count of nitrogen atoms in one Cadaverine molecule;

$27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100$

$z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2$

Molecular formula for Cadaverine:

= $C_xH_yN_z= C_{5}H_{14}N_2$

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is = $C_xH_yO_zN_w$

Count of carbon atoms in one Epinephrine molecule;

$59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100$

$x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9$

Count of hydrogen atoms in one Epinephrine molecule;

$7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100$

$y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13$

Count of oxygen atoms in one Epinephrine molecule;

$26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100$

$z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3$

Count of nitrogen atoms in one Epinephrine molecule;

$7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100$

$w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1$

Molecular formula for Epinephrine:

= $C_xH_yO_zN_w= C_9H_{13}O_3N_1$

7 0

11 days ago

At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C

VMariaS [1037]

Answer: 0.0007 moles of $CO_2$ are released when the temperature rises.

Explanation:

To determine the moles, we utilize the ideal gas law, as follows:

$PV=nRT$

where,

P = gas pressure = 1.01 bar

V = gas volume = 1L

R = gas constant = $0.08314\text{ L bar }mol^{-1}K^{-1}$

Calculated moles at T = 20° C

The gas temperature = 20° C = (273 + 20)K = 293K

Substituting values into the equation gives:

$1.01bar\times 1L=n_1\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 293K\\n_1=0.04146moles$

Calculated moles at T = 25° C

The gas temperature = 25° C = (273 + 25)K = 298K

Substituting values into the equation gives:

$1.01bar\times 1L=n_2\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 298K\\n_2=0.04076moles$

Released moles = $n_1-n_2=0.04146-0.04076=0.0007moles$

Therefore, 0.0007 moles of $CO_2$ are released when the temperature increases from 20° C to 25° C.

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13 days ago

1. Use the following thermochemical equation.

KiRa [976]

Answer:

There's a lot to address, so I'm uncertain if I can tackle this; it feels overwhelming. Perhaps you could simplify it for me as it's quite extensive.

Explanation:

I wish I had the answers.
It's too complex.
This chemistry isn't familiar to me.
If you have another chemistry-related question, feel free to ask.
This is just too difficult.

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11 days ago