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4 years ago

5

Draw the Lewis structure, including typical contributions to the resonance structure (where appropriate, allow for the possibili

ty of octet expansion, including double bonds in different positions), for (a) periodate ion; (b) hydrogen phosphate ion; (c) chloric acid; (d) arsenate ion.

1 answer:

Ber [7]4 years ago

5 0

Answer:check attached diagram

Explanation:

In order to draw a lewis structure for any ion,molecules or compounds, the following step must be followed;

FIRST STEP: select a central atom, the central atom should be the most electronegative as compare to other atoms in the ion, molecule or compounds.

For (a),(b), (c) and (d); Iodine,Phosphorus, Chlorine and Arsenic respectively.

SECOND STEP: Join this central atom to other atoms with a single bond

THIRD STEP: calculate the number of π electron using the formula below;

P= 6n +2 - V. Where V= valence electron, p= number of pie electron, n= number of atom

V can be calculated;

V= (6+6+6+6+5+1) - (-2)

=32

FOUTH STEP: complete the structure drawing it resonance

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In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (

FromTheMoon [43]

Answer:

the acceleration $a^{\to} = (0.0159 \ \ m/s^2 )i$

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So; the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

$a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}$

$a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}$

$a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}$

$a =0.0159 \ m/s^2$

Thus;

Assume the car moves in the +x direction;

the acceleration $a^{\to} = (0.0159 \ \ m/s^2 )i$

7 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

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4 years ago

Read 2 more answers

Why is it necessary for cells to be so small

madreJ [45]

Answer:

Cells are small because they need to keep a surface area to volume ratio that allows for adequate intake of nutrients while being able to excrete the cells waste.

Explanation:

That is why the cell needs to be small

8 0

3 years ago

If a yo-yo is spinning so that it makes 240 revolutions every minute, what is its period?

Aleksandr-060686 [28]

Answer: 0.25 seconds.

Explanation:

The yo-yo does 240 revolutions in one minute, and we know that one minute has 60 seconds, then the revolutions per second can be calculated as:

240 rev/60s = 4 rev/s, this will be the frequency of the yo-yo

The frequency is actually written as: f = 4 Hz = 4 s^-1

We want to find the period of this yo-yo.

The period is the duration of one cycle, and we have the relation:

f = 1/T

Where f is the frequency and T is the period, then:

T = 1/f

And we know the value of f, it is f = 4 s^-1

Then the period will be:

T = 1/(4 s^-1) = (1/4) s

Then the period of the yo-yo is 1/4 seconds = 0.25 seconds.

4 0

3 years ago

A particular engine has a power output of 8 kW and an efficiency of 37%. If the engine expels 10994 J of thermal energy in each

VARVARA [1.3K]

Answer:

(a) 17450.8 J (B) 1.239 sec

Explanation:

We have given power output =8 KW

Heat expelled = 10994 j

Efficiency =37% = 0.37

(A) We know that efficiency $\eta =1-\frac{heat\ expelled}{heat\ input}$

$0.37=1-\frac{10994}{heat\ input}$

Heat input =17450.8 J

(B) Work done by the engine is = 17450.8-10994=6456.8 j

Power output is given by 8 KW =8000 W

So time for each cycle is $\frac{8000}{6456.8}=1.239 sec$

3 0

3 years ago