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3 years ago

8

Which of the following are likely to form a covalent bond?

2 answers:

Alisiya [41]3 years ago

7 0

Two non-metals form a covalent bond.
The answer is oxygen and oxygen.

Lady bird [3.3K]3 years ago

6 0

Non-metals only makes covalent bonds and the only answer choice with no metal is b. Brainliest meee!

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If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop a

MAVERICK [17]

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current $I_1$ = 100 μA

current $I_2$ = 1 mA

forward voltage $V_r$ = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.

Using the formula:

$I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

$I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

where;

$V_r$ = 0.7

$I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}$

$I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}$

$\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

$\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

Suppose n = 1

$V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV$

Then;

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}$

${e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}$

${\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}$

${\dfrac{V_r'}{nv_T}} =29.37$

$V_r'=29.37 \times nV_T$

$V_r'=29.37 \times 25.86$

$V_r'=759.5 \ mV$

$Vr' \simeq$ 760 mV

Thus, the voltage drop across this same diode will be 760 mV

3 0

3 years ago

A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constan

san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

ρ(r) = C/r^2

Volume(V) of a sphere = 4/3(πr^3)

dV/dr = (4/3)×3πr^2

dV = 4πr^2dr

Therefore,

dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

dq = C/r^2[4πr^2dr]

dq = 4Cπdr

FOR TOTAL CHANGE 'Q', we integrate dq

∫dq = ∫4Cπdr at r = R and r = 0

∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

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3 years ago

What is an artificial situation in which a scientist deliberately makes certain changes in the world about him so that he can ob

slava [35]

I believe the process you have described is known as an 'experiment'.

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4 years ago

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Answer:

76km))))))))))))#)))

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3 years ago

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What is the gravitational potential energy of a rock with the mass of 67 kg if it is sitting on top of a hill .35 kilometers hig

solong [7]

Not to sure but maybe 23.45

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