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lutik1710 [3]
3 years ago
14

What are the series of steps that form oxygen atoms?

Chemistry
1 answer:
maw [93]3 years ago
6 0

Answer:

2.5.2 Atomistic Configurations of Oxygen in Silicon Crystals

Oxygen atoms are incorporated in the silicon lattice on an interstitial position (denoted by Oi), where the oxygen atom sits in an almost bond-centered position between two adjacent Si atoms [101]. Several of the local vibration modes are infrared (IR) active. The predominant IR line at 1104/cm (at room temperature) was calibrated with respect to other analytical methods, such as gas fusion and several radioactive techniques, and is used as the standard method for the quantitative determination of the oxygen content in as-grown silicon crystals (as-grown silicon usually contains only a negligible amount of precipitated oxygen).

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Which reaction is an example of heterogeneous catalysis?
lesya692 [45]

Answer:

Explanation:

Industrial examples

Process Reactants, Product(s)

Ammonia synthesis (Haber–Bosch process) N2 + H2, NH3

Nitric acid synthesis (Ostwald process) NH3 + O2, HNO3

Hydrogen production by Steam reforming CH4 + H2O, H2 + CO2

Ethylene oxide synthesis C2H4 + O2, C2H4O

5 0
3 years ago
What will be the ph of a buffer solution with an acid (pka6. 1) that is exactly half as concentrated as its conjugate base?
anygoal [31]

The pH of a buffer solution with acid that (PKA 6. 1) is exactly half as concentrated as its conjugate base is 6.4.

<h3>What is a buffer solution?</h3>

A buffer solution is a solution that has a maintained pH, not basic or not acidic. Its pH changes when acid or base is added to the solution.

We had to figure out the acid's concentration, which is exactly half that of its potential base.

We know that pH = pH_log

We have less than 6.1 pH so this is a conjugated base.

This will equal to 6.1 + log2 = 6.4

Thus, the pH of a buffer solution with acid is 6.4.

To learn more about buffer solutions, refer to the below link:

brainly.com/question/13169083

#SPJ4

7 0
2 years ago
What is the molarity of 15.0 g of NaBR in 0.250 L of solution
Tom [10]

n = m/M = 15/102.894 = 0.1457811 moles

c = n/V = n / 0.25 = 0.58312438 moles/dm3

6 0
4 years ago
An apprentice jeweler determines the density of a sample of pure gold to be 20.3 g/cm3. the accepted value is 19.3 g/cm3. what i
frosja888 [35]
We are tasked to find the percent error of the jeweler's density measurement through the formula:

%Error= Absolute Value(Experimental Value - Theoretical value) x 100%/ Theoretical Value

We are given by the problem,
Experimental Value=20.3 g/cm3
Theoretical Value= 19.3 g/cm3
By Substituting the details,
%Error= Absolute Value(20.3 g/cm3 - 19.3 g/cm3) x 100% / 19.3 g/cm3
%Error=Absolute Value (1)/19.3
%Error= 100/19.3
%Error= 5.18%

Therefore, 5.18% is the percent error committed by the jeweler<span />
3 0
4 years ago
Read 2 more answers
Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C. How much heat energy was released?
prohojiy [21]

Heat energy released : 167.2 kJ

<h3>Further explanation</h3>

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

m = 2000 g

c = 4.18  J/ g ° C  

∆t = 20 ° C  

\tt Q=2000\times 4.18\times 20\\\\Q=167200~J=167.2~kJ

7 0
3 years ago
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