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2 years ago

18.74 g of P4 is reacted with 9.6 g of Cl2 balanced is P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) how many PlC3 can I make?

Chemistry

1 answer:

Yanka [14]2 years ago

3 0

P4(s) + 6Cl2(g) → 4PCl3(l)

n P4 = m P4 / Mr P4

n P4 = 18.74 / 123.88

n P4 = 0.151 mol

n Cl2 = m Cl2 / Mr Cl2

n Cl2 = 9.6 / 70.9

n Cl2 = 0.135 mol

Cl2 act as limiting reactant

n PCl3 = (coef. PCl3)/(coef. Cl2) • n Cl2

n PCl3 = (4/6) • 0.135

n PCl3 = 0.09 mol

m PCl3 = n PCl3 • Mr PCl3

m PCl3 = 0.09 • 137.33

m PCl3 = 12.36 gr

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Choices for images

BabaBlast [244]

One

Let's start by stating what we know is wrong. Equilibrium is achieved when the reactants and products have a stable concentration. That makes D incorrect. Equilibrium is not established until about the 6th or 7th second.

The fact that you get any products at all means that the reactants will become products. Just who is favored has to be looked at very carefully. The products start very near 0. They go up until their concentration at equilibrium. When the reach equilibrium, the products have increased to 17. The reactants have dropped from 40 to 27. By a narrow margin, I would say the products are favored.

C is incorrect. There are still reactants left.

E is incorrect. the reactants started out with a concentration of 40. The reaction is not instantaneous. The concentration was highest at 40 or right at the beginning. This assumes that the reactants were mixed and the products were produced and the water/liquid amount has not changed.

B is incorrect. The concentration of the reactants is higher at equilibrium.

A is wrong. It is product favored.

I'm getting none of the above.

Problem Two

AgBr is insoluble (very). You'd have to work very hard to get them to separate into their elemental form. Just putting AgBr in water isn't enough. Lots of heat and lots of electricity are needed to get the elemental form.

I suppose you should pick B. Mass must be preserved. But if you balanced the equation, it would work with heat and electricity.

3 0

3 years ago

Read 2 more answers

Which of the following shows the wavelengths of light that an atom gives off when an electron falls to a lower energy level?

charle [14.2K]

The wavelengths of light that an atom gives off when an electron falls to a lower energy level corresponds to Emission spectrum , Option D is the correct answer.

<h3>What is Emission Spectrum ?</h3>

Light is absorbed or emitted when an electron jumps or falls into an energy level.

The energy of light absorbed or emitted is equal to the difference between the energy of the orbits.

Therefore , the wavelengths of light that an atom gives off when an electron falls to a lower energy level corresponds to Emission spectrum.

To know more about Emission Spectrum

brainly.com/question/13537021

#SPJ1

3 0

2 years ago

Cuántos gramos de hidróxido de sodio (NaOH) se debe de agregar a un litro de agua para obtener una disolución al 25% en masa

Tamiku [17]

Answer:

= 250 gramos

Explanation:

No ha preguntado sobre peso / peso o peso / volumen.

Para w / v:

Simplemente tome el 25% de 1000 y luego agréguelo en 1L de agua

La masa de NaOH necesaria es de 250 g.

Para w / w:

250 gramos de NaOH y 750 g de agua, hará que la solución sea 1L.

5 0

3 years ago

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

3 years ago

Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^26s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^56s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

5 0

3 years ago