Which statement describes clean air as a natural source

What is the chemistry principle for silly putty and candle

devlian [24]

Explanation:

Silly Putty -

It is an inorganic polymer , in which the molecules are linked via covalent bonding . And the bonds can easily be broken , by the application of even a small amount of pressure .

The composition of silly putty is , the Elmer's glue solution and the sodium borate in the ration , 4 : 1 .

<u>silly putty is polar , hydrophilic and binding in nature .</u>

Candle Wax -

The Wax of the candle is an ester of the ethylene glycol and fatty acids , and the chemical properties of was is ,<u> it is hydrophobic in nature and is insoluble in water as well .</u>

6 0

3 years ago

Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some

Gekata [30.6K]

Answer:

$k_{sp}=4.7 \times 10^{-13}.$

Explanation:

Now equation of tqo halves are:

Oxidation : $Ag(s)-->Ag^+(aq)+e^-$

Reduction : $AgBr(s)+e^--->Ag(s)+Br^-(aq)$

We know,

$E^o_{cell}=0.071-(0.8)=-0.729\ V.$

$\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.$

$k_{sp}=4.7 \times 10^{-13}.$

Hence, this is the required solution.

5 0

2 years ago

Select the structure that corresponds

Lostsunrise [7]

Answer:

C. both

Explanation:

8 0

2 years ago

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:

Jlenok [28]

To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles.
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6

C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!

3 0

2 years ago

What is the silver ion concentration in a solution prepared by mixing 425 mL 0.397 M silver nitrate with 427 mL 0.459 M sodium p

Lisa [10]

Answer:

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

Explanation:

$molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$

Moles of silver nitrate = n

Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)

Molarity of the silver nitrate solution = 0.397 M

$n=0.397 M\times 0.425 L=0.1687 mol$

Moles of sodium phosphate = n'

Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)

Molarity of the sodium phosphate solution = 0.459 M

$n'=0.459 M\times 0.427 L=0.1960 mol$

$3AgNO_3+Na_3PO_4\rightarrow Ag_3PO_4+3NaNO_3$

According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :

$\frac{1}{3}\times 0.1687 mol=0.05623 mol$ of sodium phosphate

This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.

So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

4 0

2 years ago