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maksim [4K]
3 years ago
8

Classify each of the following bonds as non polar covalent polar or ionic

Chemistry
1 answer:
Soloha48 [4]3 years ago
3 0
A) Polar (Cl is more electronegative than Si)
b) Nonpolar (Both atoms have the same electronegativity)
c) Ionic (Ionic bonds happen between a metal and a nonmetal)
d) Nonpolar (Hydrogen and carbon have about the same electronegativity) this is a common nonpolar bond)
You can identify the type of bon by looking at what is being bonded (nonmetal or metal) and the placement of the elements on the periodic table (electronegativity increases going up a group and going from left-right across a period).
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Why did mendeleev leave blank spaces in his version of the periodic table?.
Leviafan [203]
<h2>Answer: He left blank spaces for <em>undiscovered</em> elements to be added to the table.</h2>

Explanation:

3 0
2 years ago
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A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca
umka21 [38]

Answer:

W = 6.65 \cdot 10^{6} lbf

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

W = m\cdot g   (1)

Where m the mass and g is the gravity  

Also, we have that the mass is:

m = \rho*V  (2)    

Where ρ is the density and V the volume

We cand calculate the volume as follows:

V = L*w*d   (3)

Where L is the length, w is the wide and d is the depth  

By entering equation (2) and (3) into (1) we have:

W = \rho*L*w*d*g

W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}  

W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf              

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!

6 0
3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
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Size (length+width) approx.
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What is the basic unit of chemistry?
Luba_88 [7]
An atom hopefully this helps
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3 years ago
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