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MatroZZZ [7]
3 years ago
6

When two intermediate chemical equations are combined the same substance that appears in the same phase can be cancelled out ,pr

ovide that?
Chemistry
1 answer:
Mamont248 [21]3 years ago
5 0
Provided that the same amount of substance which is produced in the first step is then immediately consumed within the next elementary step of the reaction.
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In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigativ
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In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis. 

<span>Following are the investigative question that you can answer by doing this experiment.
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3 years ago
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What is the solute and solvent in roundup
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Answer:

solute is that we disolve in solvent

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3 years ago
What chemical reactions occur in a Nelson's cell? ​
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Explanation:

The graphite anodes are suspended into the brine. During electrolysis, Cl ions are oxidized at the anode and chlorine gas goes out of the cell, while sodium ions are reduced at the mercury cathode forming sodium amalgam. ... Hydrogen gas is obtained as a by–product at the cathode.

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1 year ago
If a shell can hold a maximum of 32 electrons, what is the value of n
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5 0
4 years ago
PH CHEM, PLEASE HELP QUICK! NO LINKS/VIRUSES PLEASE!
Contact [7]

Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH  +  HCl  →  H₂O  +  NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles

Now, we need to find the concentration of the OH⁻ ions.

[OH^{-}] = \frac{n_{NaOH}}{V}

Where V is the volume of the solution = 1.00 L                

[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L

Finally, we can calculate the pH of the solution as follows:

pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52

pH + pOH = 14

pH = 14 - pOH = 14 - 2.52 = 11.48

Therefore, the pH of the solution is 11.48.

I hope it helps you!

3 0
3 years ago
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