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Vsevolod [243]
3 years ago
9

All rocks are made up of?

Chemistry
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0

Answer:

not all rocks are made in the same type u sedamintory and lava rocks and marbel rocks but thier all made by stone

Sindrei [870]3 years ago
6 0

Answer:

The Earth is covered in a layer of solid rock called the crust. Rocks are either SEDIMENTARY , IGNEOUS, or METAMORPHIC. Almost all rocks made of minerals, but different rocks contain different mixtures of minerals.

Explanation:

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Carbonic anhydrase is a zinc-based enzyme that catalyzes the conversion of carbon dioxide to carbonic acid. In an experiment to
Contact [7]

Answer:

k=1.12x10^{-4}s

Explanation:

Hello,

By considering the rate law:

ln(\frac{[CO_2]}{[CO_2]_0} )=-kt\\

Solving for the time, we've got:

k=-\frac{ln (\frac{[CO_2]}{[CO_2]_0 })}{t}\\\\k=-\frac{ln(\frac{56.0mmol*dm^{-3} }{220mmol*dm^{-3} })}{1.22x10^{4}s }\\  \\k=1.12x10^{-4}s

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8 0
3 years ago
A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.Part A: What is the initial pH of this solution?Part B: What
Lelechka [254]

Answer:

Initial pH of this pH  = 7.453

pH after addition of 150.0 mg of HBr  =  7.35

pH after addition of 85.0 mg of NaOH 0.154  = 7.56

Explanation:

Since Ka value isn't given  

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log\frac{[Conjugate base]}{[Acid]}

pH = 7.52 + log \frac{0.15}{0.175}

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr    

             \frac{Mass}{Molar mass} \\= \frac{150 X10^{-3}g }{80.91} \\= 0.00185 }  mole

Moles of NaOCl in 100 ml buffer solution = \frac{0.15X100}{1000} = 0.015

Moles of HClO in 100 ml buffer solution = \frac{0.175X100}{1000} = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

                     

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

                                                                                     = 0.0132            

moles of HClO = 0.0175 + 0.00185

                         = 0.0194

Using Henderson equation pH = Pka + log\frac{Conjugate base}{Acid}

                                                    = 7.52 + log\frac{0.0132}{0.0194}

                                                     = 7.35

Part C

         pH after addition of 85 mg of HBr

moles of NaOH    

             \frac{Mass}{Molar mass} \\= \frac{85 X10^{-3}g }{40} \\= 0.00213 }  mole

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

                                                                                     = 0.0171            

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

                         = 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log\frac{Conjugate base}{Acid}

                                                    = 7.52 + log\frac{0.171}{0.154}

                                                     = 7.56

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Answer:

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