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alexgriva [62]
3 years ago
5

How do you express a number in scientific notation?

Chemistry
1 answer:
kykrilka [37]3 years ago
5 0

D. scientific notation is (insert decimal number) x10^(however many 0's there are)

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__________ may be shared by or transferred to other atoms.
Vsevolod [243]
D.) "Electrons" <span>may be shared by or transferred to other atoms. 

Hope this helps!</span>
8 0
3 years ago
Read 2 more answers
Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T
Ostrovityanka [42]

Explanation:

The given data is as follows.

          \Delta H = 286 kJ = 286 kJ \times \frac{1000 J}{1 kJ}

                            = 286000 J

 S_{H_{2}O} = 70 J/^{o}K,      S_{H_{2}} = 131 J/^{o}K

 S_{O_{2}} = 205 J/^{o}K

Hence, formula to calculate entropy change of the reaction is as follows.

          \Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)

                     = [(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]

                    = [(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}

                            = 286000 J - (163.5 J/K \times 298 K)

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

6 0
4 years ago
Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso
Sauron [17]

<u>Answer:</u>

<u>For A:</u> The number of arsenic atoms are 3.4\times 10^{21}

<u>For B:</u> The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

<u>Explanation:</u>

  • <u>For A:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of dimercaprol = 696 mg = 0.696 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.0056 moles of dimercaprol will contain 0.0056\times 6.022\times 10^{23}=3.4\times 10^{21} number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, 3.4\times 10^{21} number of dimercaprol molecules will bind with = 1\times 3.4\times 10^{21}=3.4\times 10^{21} number of arsenic atoms

Hence, the number of arsenic atoms are 3.4\times 10^{21}

  • <u>For B:</u>

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100     ......(1)

  • <u>For mercury:</u>

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%

  • <u>For thallium:</u>

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%

  • <u>For chromium:</u>

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0
3 years ago
1) If you have 2.6 moles of iron (III) oxide, how many molecules of iron (III)
Ugo [173]

Answer:

234

Explanation:

so 3 x 3 x 26 =234

8 0
3 years ago
How much heat is required, in calories, to raise the temperature of 57.8 g of silver from
Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

8 0
3 years ago
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