derekwadsworth.com is produced by cderekwadsworth.comnobacteriathe more quickly fossilsfound as early on as the Archaen eraStromatolites are few of the faster fossils and also are often uncovered in or constituting rocks near oceans. Castle were created by the fossilization of single-cells cderekwadsworth.comnobacteria, which developed their food with photosynthesis and still exist today. Part date ago to the Archean era. 


Which the the following includes The greatest variety of representative particles: 1 mole the water molecules, 1 mole that copper ato

derekwadsworth.com:

1 mole of salt chloride ions

Explanation:

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Gold has actually a molar massive of 197.0 g/mol. What is the concentration of yellow in seawater in moles per liter?(please show work)

derekwadsworth.com:

Since friend didnt get asked for a spefic weight of gold lets assume it to be 5g

Explanation:

5g Au/197 =.02538 mols in 1 liter the sea water is

.02538/1L which offers you a Molarity the .012538 presume 5g if they ask for a different number just replace 5 with any weight in grams



What massive of salt benzoate should be included to 160.0 ml the a 0.17 m benzoic acid solution in order to obtain a buffer through a ph

derekwadsworth.com :

The mass of sodium benzoate that have the right to be added to benzoic acid is 5.07 g

Explanation :

The trouble can be solved with the help of Henderson-Hasselbalch equation.

You are watching: Stromatolites are _____.

Step 1 : uncover concentration of salt benzoate

The equation is provided below.

*
} " alt=" p^H = p^Ka + log \frac " align="absmiddle" class="latex-formula">

We have been given,

pH = 4.30

= 0.17 M

Ka of benzoic acid can be discovered out using standard reference table i m sorry is 6.5 x 10⁻⁵

pKa = - log (Ka)

pKa = - log (6.5 x 10⁻⁵)

pKa = 4.19

Let us plug in the above values in Henderson equation.

*
}(0.17M) " alt=" 4.30 = 4.19 + log in \frac(0.17M) " align="absmiddle" class="latex-formula">

*
}0.17 " alt=" 0.11 = log \frac0.17 " align="absmiddle" class="latex-formula">

*
}0.17} " alt=" 10^0.11 = 10^log \frac0.17 " align="absmiddle" class="latex-formula">

The functions

*
cancel the end each other.

*
}0.17 " alt=" 10^0.11 = \frac0.17 " align="absmiddle" class="latex-formula">

*
}0.17 " alt=" 1.29 = \frac0.17 " align="absmiddle" class="latex-formula">

= 0.22 M

The base in this situation is salt benzoate.

Therefore concentration of salt benzoate is 0.22 M.

Step 2 : find moles of sodium benzoate

The volume of the solution is 160 mL.

Volume in l =

*

Moles of sodium benzoate deserve to be discovered as,

*

*

*

Moles of sodium benzoate are 0.0352

Step 3 : find mass of salt benzoate

Mass of salt benzoate deserve to be calculate as

Mass in grams = moles x Molar mass

The formula of sodium benzoate is C₇H₅O₂Na.

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Molar massive of sodium benzoate = 7 x 12.01 + 5 x 1.01 + 2 x 16 + 22.98

Molar mass of salt benzoate = 144.1

Mass of salt benzoate =

*

Mass of sodium benzoate = 5.07 g

5.07 grams of salt benzoate have to be added to the provided volume the benzoic acid.