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3 years ago

Which list below contains only ionic compounds?

Physics

1 answer:

garik1379 [7]3 years ago

7 0

Answer:

C. LiF, CaCl2, MgO

Explanation:

Ionic compound: It is that compound which consist of negative and position ions and held together by electrostatic force of attraction.

The ionic compound is neutral.

The ionic compound formed between metal and non-metal.

Covalent compound: It is that compound in which sharing of electrons between two atoms occur.

A. $CaCl_2,H_2O,LiF$

H2O is not ionic compound. It is polar covalent.

Therefore, it does not contain only ionic compounds.

B.CO2,K2O,P4O10

CO2 and P4O10 are not ionic .It is covalent.

Therefore, it does not contain only ionic compounds.

C. LiF,CaCl2 ,MgO

All compounds are ionic because in given compounds one ion is negative and one is metal (positive).

Therefore, it contains only ionic compounds.

D.NO2,MgO,CuCl2

NO2 is not ionic.

Therefore, it does not contain only ionic compounds.

Option C is correct.

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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

$v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_{B2}-v_{A2}$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

$2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s$

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4 years ago

The 60-mm-diameter shaft is made of 6061-T6 aluminum having an allowable shear stress of τallow = 80 MPa. Determine the maximum

grin007 [14]

The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:

Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ

As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.

5 0

4 years ago

(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas

aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = $\frac{k*q}{r}$

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = $\frac{k*q1}{x}$ + $\frac{k*q2}{(1.63m-x)}$ = 0

⇒ $k*q1* (1.63m - x) = -k*q2*x$ :

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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agasfer [191]

Answer:

b. amplitude

Explanation:

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Rufina [12.5K]

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