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4 years ago

Why do atoms bond? Describe examples of the two ways elements complete their valence

Physics

2 answers:

Murljashka [212]4 years ago

8 0

Answer:

1.) Atoms bond in order to gain more stability.

2.) They can complete their valence by sharing (taking and giving) electrons for 1.)

Explanation:

Hope This Helps

Ne4ueva [31]4 years ago

7 0

Answer: kindly see Explanation

Explanation:

Atoms bond in other to achieve a stable valence electron shell. Through bonding, atoms with incomplete valency in the outer shell may accept, lose or share electrons with another atom in other to achieve a stable outer shell or configuration.

Ionic bonding:

Ionic or electrovalent bonding simply occurs when atoms accept or lose electrons in their outer shell in other to achieve stability. Elements that require 1 atom to achieve stability accepts from elements which requires losing 1 atom in other to achieve a stable outer shell.

Example is Sodium chloride (NaCl) sodium loses one atom, which chlorine accepts and both achieve a stable outer shell.

Covalent bonding simply refers to electron sharing between atoms. Instead of losing or accepting, the atoms bond by sharing electron to achieve stability. Such is the bonding between Hydrogen and Oxygen in the formation of H2O, oxygen requires two electrons to attain stability, each hydrogen atom requires one each to attain stability, through sharing oxygen attains two more electrons, one each from the hygrogen atoms and the oxygen atom also shares one electron with each hydrogen atom. Then all the atoms atoms stability.

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A ball is thrown straight up. What can be said about the sign of work done by the force exerted by gravity while the ball moves

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3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

aivan3 [116]

Answer:

Part a)

$a = 3.65 m/s^2$

Part b)

$t = 9.5 s$

Part c)

$v_f = 55.1 m/s$

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

$v_f^2 - v_i^2 = 2ad$

$v_f^2 - 0 = 2(a)(165)$

$v_f = 125 km/h = 34.7 m/s$

now we have

$a = 3.65 m/s^2$

Part b)

Now for take off time we will have

$v_f - v_i = at$

$34.7 - 0 = 3.65 t$

$t = 9.5 s$

Part c)

$v_f = v_i + at$

$v_f = 0 + (3.65)(15.1)$

$v_f = 55.1 m/s$

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3 years ago

Un objeto que tiene como masa 80 kg acelera a razón de 10 m/s2 ¿que fuerza lo impulzo

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Answer:

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3 years ago

Read 2 more answers

An open cooking pot containing 0.5 liter of water at 20°C, 1 bar sits on a stove burner. Once the burner is turned on, the water

sesenic [268]

Answer:

a. the time required for the onset of evaporation is: 196.1 seconds and b. the time required for all of the water to evaporate is: 1328.3 seconds.

Explanation:

We need to stablish that there is 3 states at this problem. At the firts one, water is compressed liquid and the conditions for this state are: P1=100KPa,T1=20°C,V1=0.5m^3. From the compressed liquid chart and using extrapolation, we can get: v1=vf1=0.0010018 (m^3/Kg) and u1=uf1=83.95(KJ/kg). Now we can find the mass of water at the state 1 as: $m=\frac{V_{1} }{v_{1} } =\frac{0.5*10^{-3} }{0.0010018}=0.5(Kg)$ Then the liquid water is heated at a rate of 0.85KW, and its volume increase, while work is done by the system at the boundary, we can assume that the pressure remains constant throughout the entire process. At the second state the water is saturated liquid and the conditions are: P2=100KPa, T2=Tsat=99.63°C, v2=vf2=0.001043(m^3/Kg) and u2=uf2=417.36(KJ/Kg). Now we can find the work as: $W=mP(v_{2} -v_{1} )=0.5*100*(0.001043-0.0010018)=0.00207(KJ)$ . (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or $m(u_{2} -u_{1} )= Q*t-W$ , solving for t we get the time required for the onset of evaporation: $t=\frac{0.5*(417.36-83.95)+0.00207}{0.85}=196.1(s)$ .(b) Then continue heat transfer to the cooking pot and results in phase change getting vapor at 99.63°C. At the final state or third state the mass is zero because all liquid was evaporated and the initial mass at this state is the same for the second state: 0.5 (Kg) and doing an energy balances results in: $(m_{3} u_{3} -m_{2} u_{2})=Q*t-W+( m_{3}-m_{2})h_{e}$ , but m3=0, now solving for t we can get the time required for all of the water to evaporate as: $t=\frac{m_{2}(h_{e}-u_{2})+W}{Q}$ . We can get from the saturated liquid chart the enthalpy he=hge=2675.5(KJ/Kg) @P=100KPa. Now we need to calculate the work related with the volume decreases as vapor exits the control volume or process 2-3 work boundary as: $W=\int\limits^3_2 {p} \, dV= p*(V_{3} -V_{2} )=-m_{2} P_{2} v_{2} =-(0.5)*100*0.001043=-0.0522(KJ)$ . Now replacing every value in the time equation we get: $t=\frac{0.5(2675.5-417.36)+(-0.0522)}{0.85}=1328.3(s)$