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lubasha [3.4K]
3 years ago
15

What kind of intermolecular forces act between a hydrogen cyanide molecule and a chloromethane molecule? note: if there is more

than one type of intermolecular force that acts, be sure to list them all, with a comma between the name of each force?
Chemistry
1 answer:
solmaris [256]3 years ago
3 0
London dispersion forces, dipole forces
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How many atoms are in a 0.50 g sample of helium
Mrac [35]
0.5 G is your answer for this question
3 0
3 years ago
Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) 2Hg (l) + O
diamong [38]

hey there!:

2HgO (s) =>  2Hg (l) + O2 (g)

2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.

So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.

603 g of Hg = 603 / 200.6 = 3 moles

Percent yield = ( actual yield / theoretical yield) * 100

= ( 3/4) * 100

= 75 %

Hope this helps!

7 0
3 years ago
Read 2 more answers
What is the difference between a amalgam and alloy.
Gennadij [26K]
Combinations of different metals form alloys. An alloy composed by mercury and other metal (or metals) forms amalgam. When a true alloy is formed, the component metals are mixed together at a temperature which is greater than the melting point of all of them.Apr 7, 2017
8 0
4 years ago
Murcury is the only metal at room temperature. Its density is 13.6g/mL. How many grams of murcury will occupy a volume of 95.8mL
Hatshy [7]

Answer:

<h3>The answer is 1.30288 × 10³ g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 95.8mL

density = 13.6g/mL

We have

mass = 13.6 × 95.8 = 1302.88

We have the final answer as

<h3>1.30288 × 10³ g</h3>

Hope this helps you

3 0
3 years ago
A buffer is 0.282 m c6h5cooh(aq and 0.282 m na(c6h5coo(aq. calculate the ph after the addition of 0.150 moles of nitric acid to
Sindrei [870]
You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:

PH = Pka + log[acid/base]

Step 1 (Figure out the concentrations):

0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3

Step 2 (Plug into equation):

PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]

PH = 3.69

7 0
4 years ago
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