Molecules in a sample of NH3 are held closely together by intermolecular forces

Answers

Answer 1

Answer: The forces in NH3 includes the vanderwaals force which will always be present and hydrogen bonding due to the electronegativity of N as compared to H. When it is in solution, there will be ionic interactions because there will be some NH4+ ions formed.

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the relative atomic mass of boron 10.2.if the element occurs in two isotopic forms,boron-10 and boron-11 ,what are their geometrical abundances?

Answers

The correct answer for the question that is being presented above is this one:

Boron consists of two isotopes, boron-10 and boron-11.
Given that its relative atomic mass is 10.2

Find the abundance of each isotope.

Let y/100 = the abundance of copper-10
and (100 - y)/100 = the abundance of copper-11

10.2 = (y/100 x 10) + [(100 - y)/100 x 11]
10.2 = 10y/100 + 1100/100 - 11y/100
1020 = 10y + 1100 - 11y
-80 = -y
y = 80

Abundance of boron-10 = 10/100 = 10%

Abundance of boron-11 = 100 - 10 = 90%

1) This method is a technique used in separating a less-dense substance from a denser one. A. Evaporation. B. Decantation. C. Picking. D. All of the above. 2) What will happen to the less-dense substance of mixtures if you separate them using decantation method? A. Remain the same. B. It will become solid. C. It will evaporate. D. It will float up. 3) In the process of decantation, the mixture is left: A. Behind. B. Undisturbed. C. Removed. D. None of the above. 4) How can the decantation method be useful in everyday life? A. It helps us to become more productive. B. It gives us satisfaction in our everyday living. C. It's just an ordinary technique. D. It helps us to make our everyday living more efficient and easier.

Answers

Answer:

1. B Decantation. 2. D 3. B

Explanation:

This method is often used when dealing with mixtures of liquids or immiscible liquids with different densities

2. This is because the denser substance sinks to the bottom due to its higher density, while the less dense substance remains on the surface. It allows for the separation of the two substances based on their density.

3. The mixture is left undisturbed after the less dense substance has been poured off. This allows the denser substance to settle at the bottom while the lighter substance remains on top

4. It's a handy method for separating substances based on their density For example, if you have a mixture of oil and water, you can use decantation to separate the two. By letting the mixture sit undisturbed, the oil, being less dense, will float to the top, allowing you to carefully pour it off and separate it from the water.

a sample of ammonia contains 9g hydrogen and 42g nitrogen. another sample contains 5g hydrogen .calculate the amount of nitrogen in the second sample

Answers

Relative and average atomic mass both describe properties of an element related to its different isotopes. Out of these two Relative atomic mas is more accurate. Therefore, the amount of nitrogen in the second sample is 23.3g.

What is mass?

Mass defines the quantity of a substance. It is measured in gram or kilogram. Average mass is the mass of atoms of an element that are isotopes. It can be calculated by multiplying mass of a isotope to natural abundance of that isotope.

Average atomic mass = (mass of first isotope× percent abundance of first isotope)+(mass of second isotope× percent abundance of second isotope)

9 g of hydrogen - 42 g of nitrogen

5 g of hydrogen - x g of nitrogen

9x= 42 ×5

x=23.3g

Therefore, the amount of nitrogen in the second sample is 23.3g.

To learn more about mass, here:

brainly.com/question/28704035

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9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

$9x=42 \cdot 5 \n9x=210 \nx \approx 23.33$

The mass of nitrogen in the second sample is 23.33 g.

What do you mean by metallic bond?

Answers

Answer:

_________________________________________________

METALLIC BOND:

Metallic Bond is basically a electrostatic force of attraction between the metal ions which are arranged in a lattice(Lattice is a regular repeating pattern) and the free electrons floating around the metal ions.

_____________________________________

PROPERTIES OF METALLIC BOND:

1- Metals are good conductors of heat and electricity

2- Metals are ductile

3- Metals are malleable

4- Metals are solid at room temperature

5- Metals Possess Metallic luster

_____________________________________

An atom of argon in the ground state tends not to bond with an atom of a different element because the argon atom has (a)a total of two valence electrons (b)more protons than neutrons (c)more neutrons than protons or (d)a total of eight valence electrons

Answers

Answer: Option (d) is the correct answer.

Explanation:

Atomic number of argon is 18 and its electronic configuration is $1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)$ .

Also, it is known that number of electrons in K, L and M shells of argon are 2, 8, 8.

Thus, we can conclude that an atom of argon in the ground state tends not to bond with an atom of a different element because the argon atom has a total of eight valence electrons.

So, as per the octet rule, last sub-shell of argon is completely filled. Hence, it will not form any bond with any other atom.

the answer is a total of eight valence electrons
hope this help

CH3COOH mc015-1.jpg CH3COO– (aq) + H+(aq)What will happen to the chemical equilibrium of the solution if CH3COONa is added?

Answers

The equilibrium will shift to the left or the backward reaction since addition of CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted since the reactions tends to go to a new equilibrium because of the addition.

Answer:

The equilibrium will shift to the reactants and increase the concentration of CH3COOH

Explanation:

According to Le Chatelier's principle when a system at equilibrium is subjected to a change in temperature, pressure, concentration etc then the equilibrium will shift in a direction to undo the effect of the induced change.

The given reaction is:

CH3COOH ↔ CH3COO-(aq) + H+(aq)

CH3COONa exists as ions i.e. CH3COO- and Na+. Therefore, addition of CH3COONa will introduce more CH3COO- ions into the system as a result the equilibrium will shift to the left i.e. towards the reactants and will produce more of CH3COOH

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