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emmainna [20.7K]

3 years ago

6

The boiling points of the elements helium, neon, argon, krypton, and xenon increase in that order. Which of the following statem

ents accounts for this increase?
1. The London dispersion forces increase.
2. The hydrogen bonding increases.
3. The dipole-dipole forces increase.
4. The dipole-induced dipole forces increases.

1 answer:

blagie [28]3 years ago

4 0

Answer:

The boiling point of noble gas increases as you go down the group because the London dispersion forces increase.

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4 years ago

How many liters of hydrogen gas is formed from the complete reaction of 15.2 g C? Assume that the hydrogen gas is collected at a

ira [324]

Answer:

38 L

Explanation:

There is some info missing. I think this is the original question.

<em>Consider the chemical reaction: C(s) + H₂ O(g) ⟶ CO(g) + H₂ (g). How many liters of hydrogen gas is formed from the complete reaction of 15.2 g C? Assume that the hydrogen gas is collected at a pressure of 1.0 atm and a temperature of 360 K.</em>

<em />

Step 1: Write the balanced equation

C(s) + H₂ O(g) ⟶ CO(g) + H₂ (g)

Step 2: Calculate the moles corresponding to 15.2 g of C

The molar mass of C is 12.01 g/mol.

$15.2g \times \frac{1mol}{12.01g} = 1.27 mol$

Step 3: Calculate the moles of H₂ produced from 1.27 moles of C

The molar ratio of H₂ to C is 1:1. The moles of H₂ produced are 1/1 × 1.27 mol = 1.27 mol.

Step 4: Calculate the volume of H₂

We will use the ideal gas equation.

$P \times V = n \times R \times T\\V = \frac{n \times R \times T}{P} = \frac{1.27mol \times \frac{0.0821atm.L}{mol.K} \times 360K}{1.0atm}= 38 L$

3 0

4 years ago

Determine the concentration of nh3(aq) that is required to dissolve 743 mg of agcl(s) in 100.0 ml of solution. the ksp of agcl i

AnnyKZ [126]

Answer:

1.1 M

Explanation:

The dissociation of $AgCl_{(s)}$ is as follows:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

Given Value for $K_{sp} = 1.77*10^{-10}$

The equation for the reaction for the formation of complex ion $Ag(NH_3)^+_2$ is :

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

The value of $K_f = 1.6*10^7$

If we combine both equation and find the overall equilibrium constant will be:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

<u> </u>

$AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

$K = (1.77*10^{-10})(1.6*10^7)$

$K = 0.00283$

If $[NH_3]$ = x M

The solubility of $AgCl_{(s)}$ in the $NH_3$ solution will be:

$x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}$

$x = 0.0518 M$

Constructing an ICE Table; we have :

$AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

Initial (M) x 0 0

Change (M) -2 (0.0518) + 0.0518 + 0.0518

Equilibrium (M) x - 0.1156 0.0518 0.0518

Equilibrium constant;

(K) = $\frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}$

$0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}$

$0.00283 = (\frac{0.0518}{x-0.1156})^2$

$x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }$

x = [NH₃] = 1.089 M

[NH₃] ≅ 1.1 M

8 0

3 years ago

How many moles is 2.04x1023 molecules of NH3?

tamaranim1 [39]

Answer:

We will Multiply the number of moles with Avagadro's number

So it will be (2.04×1023)×6.022×10^23

<h3>or 2.04×1023 NA</h3>

NA means Avagadro's number

5 0

3 years ago