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Naya [18.7K]
3 years ago
6

Of the following substances, only __________ has London dispersion forces as its only intermolecular force. CH3OH NH3 H2S CH4 HC

l
Chemistry
1 answer:
QveST [7]3 years ago
7 0

Answer:

CH₄

Explanation:

CH₃OH has hydrogen bonding due to the OH group present

NH₃ also has hydrogen bonding due to the NH bonds

H₂S has dipole-dipole forces present due to the polar SH bonds

HCl also has dipole-dipole forces due to the polar HCl bond

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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
How many grams are in 1.23 x 1020 atoms of arsenic?
Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0
2 years ago
Chalk markings cannot be easily rubbed off from a black board if kept for a long time.why?
snow_tiger [21]

The chalk particles embed themselves into the small pores on the surface.

Although a chalkboard seems smooth to the touch, it is quite rough at the microscopic level, with <em>pores</em> that reach below the surface.

When you drag chalk across the board, friction causes small particles of chalk to rub off onto the surface.

If you leave the markings for a long time, some of the chalk particles will work their way into the pores.

A brush will remove the surface particles, but <em>it will not be able to get at the particles in the pores</em>.

3 0
3 years ago
How do tidal bulges form?
Troyanec [42]

Answer: The pull of the moon's gravity on Earth's water causes tidal bulges to form on the side closest to the moon and farthest from the moon. In the place where there are tidal bulges, high tide occurs along coastline.

Explanation:

4 0
2 years ago
It takes 281.7 kJ of energy to remove 1 mole of electrons from the atoms on the surface of lithium metal. If lithium metal is ir
Alex17521 [72]

Answer:

The maximum kinetic energy of electron is = 2.93 × 10^{-19} Joule

Explanation:

We know that total energy

E = \frac{hc}{\lambda}  ------------ (1)

Here h = plank's constant = 6.62 × 10^{-34} J s

c = speed of light = 3 × 10^{8} \frac{m}{s}

\lambda = 261 nm = 261  × 10^{-9} m

Put all these values in equation (1) we get

E = 7.6   × 10^{-19} J

We know that

Total energy = Energy to remove an electron + K.E of electron

Energy to remove an electron = \frac{281.7 (1000)}{(6.023)10^{23} }

Energy to remove an electron = 4.67  × 10^{-19} J

K.E of electron = Total energy - Energy to remove an electron

K.E of electron = 7.6   × 10^{-19} -  4.67  × 10^{-19}

K.E of electron = 2.93 × 10^{-19} Joule

Therefore the maximum kinetic energy of electron is = 2.93 × 10^{-19} Joule

3 0
3 years ago
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