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3 years ago

How does an atom behave if it has a full octet of valence electrons?

Chemistry

2 answers:

erastova [34]3 years ago

6 0

Answer:

A) Very stable.

Explanation:

A valence electron ring takes up to 8 valence electrons. If it has full rings, it is stable.

Naya [18.7K]3 years ago

4 0

A) very stable this is ur anwswer

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The northern lights and huge power surges in technology are both effects of what

umka2103 [35]

Solar winds. Plasma particles from the sun affect earth's atmosphere and cause magnetic interferance.

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3 years ago

A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr

sergey [27]

Answer:

184.62 ml

Explanation:

Let $p_1, v_1,$ and $T_1$ be the initial and $p_2, v_2,$ and $T_2$ be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

$p_1=p_2$ ...(i)

$v_1$ = 200 ml

$T_1= 26 ^{\circ}C = 273+26 =299$ K

$T_2= 3 ^{\circ}C = 273+3 =276$ K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

$p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)$

For the final condition,

$p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)$

Equating equation (i), and (ii)

$\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}$

$\frac{v_1}{T_1}=\frac{v_2}{T_2}$ [from equation (i)]

$v_2=\frac{T_2}{T_1} \times v_1$

Putting all the given values, we have

$v_2=\frac{276}{299} \times 200 = 184.62 \; ml$

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0

3 years ago

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0

3 years ago

Read 2 more answers

PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE H

vfiekz [6]

Answer:

AU9NJ-BLVHV-TCLJS-54YTB

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3 years ago

Answer to the ? In letter b

andreev551 [17]

The answer to letter b is

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2 years ago