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3 years ago

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Consider the gas phase reaction 4HCl + O2 → 2Cl2 + 2H2O. What volume of chlorine gas at STP can be prepared from the reaction of

600. mL of gaseous HCl, measured at STP, with excess oxygen?
a. 150 mL b. 267 mL c. 300 mL d. 425 mL e. 600 mL

1 answer:

SOVA2 [1]3 years ago

7 0

Answer:

c

Explanation:

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Find the pressure of a gas if the volume is 2.00 L, the temperature is 310 Kelvin and

kipiarov [429]

Answer:

25.42 atm

Explanation:

Data Given:

Volume of a gas ( V )= 2.00 L

temperature of a gas ( T ) = 310 K

number of moles (n) = 2 mol

Pressure of a gas ( P ) = to be find

Solution:

Formula to be used

PV= nRT

Rearrange the above formula

P = nRT / V . . . . . . . . . . (1)

Where R is ideal gas constant

R = 0.08205 L atm mol⁻¹ K⁻¹

Put values in equation 1

P = nRT / V

P = 2 mol x 0.08205 L atm mol⁻¹ K⁻¹ x 310 k / 2 L

P = 50.84 L atm / 2 L

P = 25.42 atm

P ressure of gas (P) will be = 25.42 atm

7 0

3 years ago

An ionic bond occurs between what particles

kakasveta [241]

<span>The correct answer is that an ionic bond forms between charged particles. To form this bond, the particles transfer valence electrons (those in the outermost orbit). Specifically, in ionic bonding, the metal atom loses its electrons (thus becoming positive) and the nonmetal atom gains electrons (thus becoming negative).</span>

4 0

3 years ago

Read 2 more answers

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$

By knowing the moles of gold:

$n_{Au}=100g*\frac{1mol}{197g} =0.508mol$

It is possible to write the aforementioned formula in terms of the variable $x$ representing the moles of silver:

$20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]$

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

$x=n_{Ag}=21.0molAg$

So the mass is:

$m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g$

Best regards!

3 0

3 years ago

Which climate is most likely found in a high pressure belt

Mice21 [21]

Answer:subtropical highs. ... Near the poles the pressure is high and it is known as the polar high. These pressure belts are not permanent in nature

Explanation: The horse latitudes are subtropical regions known for calm winds and little precipitation. ... Unable to sail and resupply due to lack of wind, crews often ran out of drinking water. To conserve scarce water, sailors on these ships would sometimes throw the horses they were transporting overboard.

Hope this was Helpful

5 0

3 years ago

Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital

Thepotemich [5.8K]

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a <u>completely filled 5f, 6d, 7s and 7p orbitals</u>.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the <u>number of electrons</u> = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = <u>118</u>.

<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

7 0

3 years ago