Using isotope, can we trace back again our ancestors? How?

Chemistry

1 answer:

lidiya [134]3 years ago

4 0

Answer:

<h2><em><u>The answer is yes, we can trace back again our ancestors by using isotope. These isotopes are not just there to just exist, some of them are also useful in different kind of field especially in history and medical</u></em>.</h2>

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An example of a double replacement chemical reaction takes place in batteries. In an alkaline battery, Manganese oxide reacts wi

ozzi

Answer:

2MnO2 + H2O => Mn2O3 + 2OH is the correct balanced equation.

6 0

3 years ago

Calculate the mass of Octane needed to release 6.20 mol Co2

n200080 [17]

The combustion reaction of octane is as follow,

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O

According to balance equation,

8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted

So,

6.20 moles of CO₂ will release when = X g of Octane is reacted

Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol

X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.

8 0

3 years ago

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$