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3 years ago

Which of the following best defines conductivity?

Chemistry

1 answer:

astraxan [27]3 years ago

6 0

B. ability of a substance to transfer energy 
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astraxan [27]

Answer/Explanation:

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Kavinsky

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2 years ago

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What are the functions of protein in organisms?

Juliette [100K]

Proteins make up many many important structures in our cells ( ex : receptors , Enzymes , integral proteins and so on )

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3 years ago

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

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3 years ago

A sulfur ion has 16 protons and 18 electrons. Find the net charge

andrezito [222]

I think its 2-, it gained 2 electrons.

5 0

3 years ago

A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base

Gekata [30.6K]

When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
= 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial 0.00525 0 0
change - 0.002625 +0.002625 +0.002625
equilibrium 0.002625 0.002625 0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
= -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
= 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74

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3 years ago

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