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3 years ago

I need help. ASAP! Look at the picture below. (4)

Chemistry

2 answers:

Bezzdna [24]3 years ago

7 0

The answer would be B, double covalent bond... I would think.

jekas [21]3 years ago

3 0

I believe the answer is B.

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Generally, as you go from left to right on the periodic table, what happens to:

Stolb23 [73]

Answer:

A.Atomic Radius I heard something similar when my teacher explained this to my class

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2 years ago

What properties of a substance determine how that substance will react when combined with other substances

Semmy [17]

Answer:

Chemical Properties

Explanation:

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2 years ago

This infectious disease requires a special mask for protection

11111nata11111 [884]

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3 years ago

Read 2 more answers

Does standard enthalpy use net ionic equations?a. Trueb. False

Ad libitum [116K]

Answer:

True

Explanation:

In an ionic reaction, the net ionic equation is used to calculate the standard enthalpy of reaction.

Hence the standard enthalpy of any ionic reaction can be obtained by considering its balanced net ionic reaction equation, hence the answer above.

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2 years ago

A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/

Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of high-pressure side is $C_1 = 0.722 \ kg/m^3$

Explanation:

From the question we are told that

The thickness of the polyethylene is $d = 1.5 \ mm = 0.0015 \ m$

The temperature is $T = 600 \ K$

The flux is $JA = 2.48 *10^{-5} \ kg/m^2\cdot s$

The concentration on the low-pressure side is $C_2 = 0.5 \ kg/m^3$

The initial diffusivity is $D_o = 6.2 *10^{-4} \ m^2 /s$

The activation energy for diffusion is $Q_d = 41 \ kJ /mol = 41*10^3 J /mol$

Generally the diffusivity of the oxygen at 600 K can be mathematically evaluated as

$D = D_o * e^{- \frac{Q_d}{R * T } }$

substituting values

$D = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600 } }$

$D = 1.671 *10^{-7} \ m^2 /s$

Generally the flux is mathematically represented as

$JA = D * \frac{C_1 -C_2}{d}$

Where $C_1$ is the concentration of oxygen at the higher side

$C_1 = d * \frac{JA}{D} + C_2$

substituting values

$C_1 = 0.0015 * \frac{2.48*10^{-5}}{1.671*10^{-7}} + 0.5$

$C_1 = 0.722 \ kg/m^3$

6 0

3 years ago