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2 years ago

6. List the following particles in size order, from biggest to smallest?

Chemistry

1 answer:

Arte-miy333 [17]2 years ago

5 0

Answer:

the particles, molecules, and gas range from smallest to largest in the following order: electron < proton < atom <...

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What type of bond exists between the atoms in a molecule of N2?

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2 years ago

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How much sunlight is there in the transition zone. For science.

soldi70 [24.7K]

Answer:

Transition zone of the sun

Explanation:

Transition Region - The transition region is a very narrow (60 miles / 100 km) layer between the chromosphere and the corona where the temperature rises abruptly from about 8000 to about 500,000 K (14,000 to 900,000 degrees F, 7700 to 500,000 degrees C).

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2 years ago

An unknown metal displaces cadmium (Cd) from solution but does not displace chromium (Cr). Using the activity series, determine

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This is the order of activity in decreasing way for the mentioned metals

Mn>Zn>Cr>Fe>Cd>Co>Ni>Pb

Fe is in between Cr and Cd, it is more active than Cd and less active than Cr, so it will displace Cd but not Cr.

Answer:Fe (it will displace Cd, but not Cr)

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3 years ago

This is a summary of the findings of an experiment

olga nikolaevna [1]

Conclusion is the answer you are looking for

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3 years ago

If 12.8 g lead(II) sulfate (303.3 g/mol) precipitates when excess potassium chloride is added to 1.65 L of a water sample, what

quester [9]

Answer:

$M=0.0256M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$PbSO_4(aq)+2KCl(aq)\rightarrow PbCl_2(s)+K_2SO_4(aq)$

In such a way, since all the lead (II) is converted due to the excess of potassium chloride, the moles of lead (II) in the sample are computed from the mass of lead (II) sulfate:

$n_{Pb^{2+}}=12.8gPbSO_4*\frac{1molPbSO_4}{303.3gPbSO_4} *\frac{1molPb^{2+}}{1molPbSO_4} \\\\n_{Pb^{2+}}=0.0422molPb^{2+}$

Thus, since volume of the solution is 1.65 due to the fact that the addition of the reactants is not enough to significantly modify the reaction's volume, the resulting molar concentration of the lead (II) ions is:

$M=\frac{n_{Pb^{2+}}}{V}=\frac{0.0422molPb^{2+}}{1.65L}\\ \\M=0.0256M$

Regards.

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3 years ago