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1 month ago

The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um

Chemistry

1 answer:

lions [2.9K]1 month ago

6 0

$\boxed{\sf 1Å=10^{-10}m}$

$\\ \rm\longmapsto 2.27Å$

$\\ \rm\longmapsto 2.27\times 10^{-10}m$

$\\ \rm\longmapsto 0.227\times 10^{-9}m$

$\\ \rm\longmapsto 0.0227\times 10^{-8}m$

$\\ \rm\longmapsto 0.00227\times 10^{-7}m$

$\\ \rm\longmapsto 0.00023\times 10^{-6}m$

$\\ \rm\longmapsto 0.00023\mu m$

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How many electrons are involved in one equivalent of oxidation-reduction?

Alekssandra [3086]

One electron is involved. Explanation: In redox reactions, determining the equivalents requires knowledge of the number of transferred electrons. In this specific case, one equivalent corresponds to a transfer of a single electron.

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29 days ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

lions [2927]

Answer: The molecular formula for the specified organic compound is $C_{18}H_{20}O_2$

Explanation:

The combustion reaction of a hydrocarbon comprising carbon, hydrogen, and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' denote the subscripts of Carbon, hydrogen, and oxygen respectively.

The information provided includes:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

From our knowledge:

Molar mass of carbon dioxide is 44 g/mol

Molar mass of water is 18 g/mol

For determining the amount of carbon:

In carbon dioxide weighing 44 g, 12 g of carbon is found.

Hence, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ grams of carbon will be found.

For finding the mass of hydrogen:

In water weighing 18 g, 2 g of hydrogen can be found.

Thus, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ grams of hydrogen will be present.

Mass of oxygen in the compound is given by (13.42) - (10.80 + 1.00) = 1.62 g

To derive the empirical formula, the following steps must be followed:

Step 1: Convert the indicated masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the ratio of moles of the respective elements.

To find the mole ratio, each mole value is divided by the smallest amount of moles calculated, which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3:Using the mole ratios as subscripts.

The ratio of C: H: O = 9: 10: 1

Therefore, the empirical formula for the mentioned compound is $C_9H_{10}O$

To ascertain the molecular formula, it is necessary to find the valency, which is multiplied by each element to derive the molecular formula.

The equation to determine the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

Given the data:

Molecular formula mass = 268.34 g/mol

Empirical formula mass = 134 g/mol

Substituting the values into the aforementioned equation yields:

$n=\frac{268.34g/mol}{134g/mol}=2$

By multiplying this valency with each element's subscripts from the empirical formula, the results are:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Consequently, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

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21 day ago

Part 1 Designing an Investigation

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20 days ago

A balloon filled with 1.22 L of gas at 286 K is heated until the

VMariaS [2998]

Response: 670K

Rationale:

Provided data includes:

Initial volume of gas V1 = 1.22 L

Initial temperature T1 = 286 K

Final volume V2 = 2.86 L

Final temperature T2 =?

As temperature and volume are interrelated when pressure remains constant, apply Charles' law: