When working with scientific data, the average of a data set is called the

Chemistry

1 answer:

sasho [114]2 years ago

3 0

Answer:

Its called the mean of the data.

Explanation:

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How many molecules of Oxygen are consumed if 24 molecules of Carbon Dioxide are produced?

Nat2105 [25]

Answer:

Explanation:

C + O2 → CO2

Mole of C = 24 g/(12 g/mole)

Mole of C = 2 mole

Mole of molecular O2 = 74 g/(32 g/mole)

Mole of molecular O2 = 2.3125 mole

Since mole of C < mole of O2, then C being the limiting reagent.

From the reaction, it shows that mole ratio between C and O2 = 1 : 1.

So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.

Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.

Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.

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2 years ago

Why do atomic model has changed over time

Elenna [48]

The model of the atom has dramatically changed over many many years.We learned atoms make up different substances and are the smallest particles of matter, which have subatomic particles that are very small portions of matter. At first scientist only thought there were electrons which are negatively charged.

8 0

2 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$