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3 years ago

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Which of the following state of matter has the modt space between the particles

1 answer:

lesya [120]3 years ago

8 0

Answers

(A)Flammability

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I need help with number 1 :/ please help me, thank you

Elena L [17]

Answer:

Amplitude is measured from the center line to the highest point in the waves.

Wavelength is the distance between one wave to the other from the highest point.

Frequency is the rate of the waves.

7 0

3 years ago

A neutralization reaction only produces a salt when a strong acid reacts with a strong base.

Svetach [21]

B. False (makes water as well)

7 0

3 years ago

A voltaic cell is constructed with two silver-silver chloride electrodes, where the halfreaction is AgCl (s) + e- → Ag (s) + Cl-

densk [106]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-$

<u>Reduction half reaction (cathode):</u> $AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}]_{diluted}$ = 0.0222 M

$[Cl^{-}]_{concentrated}$ = 2.22 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

6 0

3 years ago

The mole fraction of he in a gaseous solution prepared from 4.0 g of he, 6.5 g of ar, and 10.0 g of ne is __________.

solniwko [45]

Given that there is 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne :

So, first all these masses are converted into moles:

$Numer of moles= \frac{Given mass}{Molar mass}$

$4.0 g of He =\frac{4 g}{4 \frac{g}{mol}}=1 mol$

$6.5 g of Ar =\frac{6.5 g}{39.95 \frac{g}{mol}} = 0.1627 mol$

$10.0 g of Ne =\frac{10 g}{20.18 \frac{g}{mol}} = 0.4955 mol$

Total Number of moles = 1 + 0.1627 + 0.4955 = 1.658 mol

$Mole fraction of He in substance =\frac{Number of moles of He }{Total number of moles}$

= $\frac{1}{1.695}=0.6$

3 0

3 years ago

Which has the greatest mass? 1 mole Ca 1 mole CO 1 mole O2 1 mole CH4 1 mole NO

tino4ka555 [31]

Answer: 1 mole of $Ca$ has the greatest mass.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP , contains avogadro's number $(6.023\times 10^{23})$ of particles and weighs equal to the molecular mass of the substance.

1 mole of $Ca$ has a mass of 40 g.

1 mole of $CO$ has a mass of 28 g

1 mole of $O_2$ has a mass of 32 g

1 mole of $CH_4$ has a mass of 16 g.

1 mole of $NO$ has a mass of 30 g.

Thus the greatest mass is of 1 mole of $Ca$

8 0

3 years ago