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2 years ago

A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?

Chemistry

1 answer:

REY [17]2 years ago

8 0

Answer:

$Sn_2O$

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

$\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%$

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

$n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO$

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

$Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1$

Therefore, the empirical formula is:

$Sn_2O$

Best regards.

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Answer:

0.129g MgCl2

Explanation:

For this we need to understand the concept of molarity.

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M=n/L

Here we are given molarity of 0.054M and volume of 25ml. we just plug this in formula to find moles of MgCl2

0.054=x/(25/1000) (we divided 25 by 1000 to convert it to litres of solution)

x=0.00135 moles of MgCl2 (we are not done yet the question asks for grams so to convert to grams we multiply by molar mass of MgCl2.)

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2 years ago

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$Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)$

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

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233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

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2 years ago

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Neither. A basic postulate of the Kinetic Molecular Theory is that the molecules of <em>all gases</em> at the same temperature have the same average kinetic energy.

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Anika [276]

Answer:

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Our compound is the RbOH.

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As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

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2 years ago

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Anestetic [448]

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