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4 years ago

Given the same conditions of temperature and pressure, which noble gas will diffuse most rapidly

2 answers:

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<span>Given the same conditions of temperature and pressure the noble gas He will diffuse most rapidly. </span>He is the smallest noble gas and so it can diffuse (pass through) membranes the easiest.It is colorless, odorless, tasteless, non-toxic chemical element.
Correct answer:A

Natasha2012 [34]4 years ago

0 0

<span>At the same conditions of temperature and pressure the noble gas with the smallest density will diffuse faster. But since density is proportional to the molecular mass. A look at the periodic table shows that Helium will diffuse faster. The correct answer is A.</span>

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molecular orbital theory correctly predicts paramagnetism according to mo theory, overlap of two p atomic orbitals produces .oxy

Mkey [24]

Overlap of two p atomic orbital produces oxygen gas ,o2 because it has two unpaired electron and explains paramagnetism.

O2 molecules have two unpaired electrons Materials in which all the electrons are paired are diamagnetic. It repel a magnetic field.Para magnetic and diamagnetic materials do not act as permanent magnets. Molecular orbital theory provides an explanation of chemical bond that explains paramagnetism of the oxygen molecule. It also explains the bonding in a other molecule. O2 molecule has two unpaired electron. Electrons creates magnetic field as it goes to the ground state and also spin.MO theory describes the distribution of electrons in molecules same as the distribution of electrons in atoms described using atomic orbital.

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1 year ago

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

Maksim231197 [3]

The theoretical yield of $SnS_2$ will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of $SnBr_4$ to $Na_2S$ is 1:2.

Mole of 48.1 mL, 0.478 M $SnBr_4$ = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M $Na_2S$ = 0.160 x 48.8/1000 = 0.0078 moles

$SnBr_4$ $Na_2S$ is the limiting reactant.

Mole ratio of $SnBr_4$ and $SnS_2$ = 1:1

Equivalent mole of $SnS_2$ = 0.023 moles

Mass of 0.023 noles $SnS_2$ = 0.023 x 182.81 = 4.20 grams

With 0.0333 g of $SnS_2$ recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

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Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

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2 years ago

Which is the electron configuration for zinc?

FinnZ [79.3K]

Answer:

1s2 2s2 2p6 3s2 3p6 4s2 3d10

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2 years ago

Group 1 elements have an average electronegativity of 0.84 (not

sergey [27]

A) The bond that will be formed by these elements is : <u>Ionic Bond </u>

B) Group 1 elements Gives up its electron to Group 17 elements

<h3><u /></h3><h3><u>Electronegativity values </u></h3>

From the electronegativity values given in the question group 17 elements with an electronegativity value of 2.99 are more electronegative than group 1 elements with a value of 0.84. Therefore they accept electron transfer from group 1 elements.

The acceptance of electrons from group 17 elements leads to formation of an ionic bond between group 1 and group 17.

Hence we can conclude that the bond that will be formed by these elements is Ionic bond and Group 1 elements gives up its electron to group 17 elements.

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2 years ago

What is the ratio by mass of oxygen to carbon when 32 g of oxygen combine with 12 g of carbon?

Mrac [35]

CO_2
1:2
32gO(1molO/16gO)=2molO
12gC(1molC/12gC)=1molC

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4 years ago