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2 years ago

13

The mass of radium-226 in a sample is found to have decreased from 45.00g to 5.625g in a period of 4800 years.From this informat

ion. Calculate the half life of radium-226

1 answer:

yan [13]2 years ago

4 0

Answer:

Half life = 1600 years

Explanation:

Given data:

Total mass of sample = 45.00 g

Mass remain = 5.625 g

Time period = 4800 years

Half life of radium-226 = ?

Solution:

First of all we will calculate the number of half lives passes,

At time zero 45.00 g

At first half life = 45.00 g/ 2= 22.5 g

At 2nd half life = 22.5 g/ 2 = 11.25 g

At 3rd half life = 11.25 g/ 2= 5.625 g

Half life:

Half life = Time elapsed / number of half lives

Half life = 4800 years / 3

Half life = 1600 years

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Functional group and bond hybridization of vanillin

lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.

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2 years ago

Element z reacts with bromine to form an ionic compound zbr2. A) state with reasons i) the physical properties of zbr2 at room t

kondor19780726 [428]

Answer:

The answer to your question is:

Explanation:

A) state with reasons

From the description we can conclude that the compound is a binary salt. Binary salts are form by a metal and a non metal, one of their characteristics is that the elements are attached by an ionic bond.

i) the physical properties of ZBr2 at room temperature

Crystals, high fusion and boiling points, they conduct water when dissolved in water, they are soluble in water but not in gasoline.

ii) whether z is a metal or non metal.

Z is a metal

3 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as $C_7 H_6 O_3$ </em>

<em></em>

<u>Explanation:</u>

$Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$$

$mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

$moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

$\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$$

So the empirical formula is $C_7 H_6 O_3$

Empirical formula mass

$=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g$

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

$=1 \times C_7 H_6 O_3$

Compound X = $C_7 H_6 O_3$ is the Answer

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2 years ago

Which of the following statements will decrease the amount of work the system could perform? (a) Add to the solution (b) Add sol

Dmitry_Shevchenko [17]

Answer:

Increase the concentration of the solution

Explanation:

Increasing the solute would increase the concentration. Increasing the solvent would decrease the concentration. For instance, if your lemonade was too tart, you would add more water to decrease the concentration

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2 years ago