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Lerok [7]
2 years ago
5

Friction. That was the topic Sally and Sue are investigating in science. They used a small car and a ramp as seen in the

Chemistry
1 answer:
DaniilM [7]2 years ago
3 0

Answer: A: The smoother the ramp surface, the greater the car's speed.

Explanation: I did the USA test prep

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Based upon the following diagram, propose a possible identity for atoms X and Y. Explain your answer in terms of the periodic ta
zhenek [66]

Answer:

Up until now we have been discussing only the elemental forms of atoms which are neutrally charged. This is because the number of electrons (negative in charge) is equal to the number of protons (positive in charge). The overall charge on the atom is zero, because the magnitude of the negative charge is the same as the magnitude of the positive charge. This one-to-one ratio of charges is not, however, the most common state for many elements. Deviations from this ratio result in charged particles called ions.

Throughout nature, things that are high in energy tend to move toward lower energy states. Lower energy configurations are more stable, so things are naturally drawn toward them. For atoms, these lower energy states are represented by the noble gas elements. These elements have electron configurations characterized by full s and p subshells. This makes them stable and unreactive. They are already at a low energy state, so they tend to stay as they are.

The elements in the other groups have subshells that are not full, so they are unstable when compared to the noble gases. This instability drives them toward the lower energy states represented by the noble gases that are nearby in the periodic table. In these lower energy states, the outermost energy level has eight electrons (an “octet”). The tendency of an atom toward a configuration in which it possesses eight valence electrons is referred to as the “Octet Rule.”

There are two ways for an atom that does not have an octet of valence electrons to obtain an octet in its outer shell. One way is the transfer of electrons between two atoms until both atoms have octets. Because some atoms will lose electrons and some atoms will gain electrons, there is no overall change in the number of electrons, but with the transfer of electrons the individual atoms acquire a nonzero electric charge. Those that lose electrons become positively charged, and those that gain electrons become negatively charged. Recall that atoms carrying positive or negative charges are called ions. If an atom has gained one or more electrons, it is negatively charged and is called an anion. If an atom has lost one or more electrons, it is positively charged and is called a cation. Because opposite charges attract (while like charges repel), these oppositely charged ions attract each other, forming ionic bonds. The resulting compounds are called ionic compounds.

The second way for an atom to obtain an octet of electrons is by sharing electrons with another atom. These shared electrons simultaneously occupy the outermost shell of both atoms. The bond made by electron sharing is called a covalent bond. Covalent bonding and covalent compounds will be discussed in Chapter 4 “Covalent Bonding and Simple Molecular Compounds”.

At the end of chapter 2, we learned how to draw the electron dot symbols to represent the valence electrons for each of the elemental families.  This skill will be instrumental in learning about ions and ionic bonding. Looking at Figure 3.1, observe the Noble Gas family of elements. The electron dot symbol for the Nobel Gas family clearly indicates that the valence electron shell is completely full with an octet of electrons.  If you look at the other families, you can see how many electrons they will need to gain or lose to reach the octet state.  Above, we noted that elements are the most stable when they can reach the octet state. However, it should also be noted that housing excessively high negative or positive charge is unfavorable.  Thus, elements will reach the octet state and also maintain the lowest charge possible.   You will note that for the IA, IIA, IIIA and transition metals groups, it is more economical to lose electrons (1-3 electrons) from their valence shells to reach the octet state, rather than to gain 5-7 electrons.  Similarly main group columns VA, VIA, and VIIA tend to gain electrons (1-3) to complete their octet, rather than losing 5-7 electrons. Some atoms, like carbon, are directly in the middle.  These atoms don’t like to gain or lose electrons, but tend to favor the sharing model of chemical bonding. The remaining sections of this chapter will focus on the formation of ions and the resulting ionic compounds.

Explanation:

8 0
2 years ago
In an open system the vapor pressure is equal to the
erica [24]

Answer: In an open system the vapor pressure is equal to the outside air pressure.

Explanation:

An open system is a system that is capable of exchanging both matter and energy with its surroundings.

For example, a hot cup of coffee is an open system as the cup is not covered with a lid. Therefore, heat energy from the coffee goes into the surroundings.

Vapor pressure is defined as the ability of a substance to change into vapor state. pressure

Therefore, we can conclude that in an open system the vapor pressure is equal to the outside air pressure.

5 0
2 years ago
7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are
Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] =  0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0
3 years ago
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Apvartanank ka sutra in hindi<br>​
lana [24]

Answer:

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What does anhydrous mean in your own words?
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Answer: no water

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