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Arlecino [84]
3 years ago
14

What is the H* concentration in a solution with a pH of 1.25?​

Chemistry
1 answer:
Kamila [148]3 years ago
7 0

Answer:

.056

Explanation:

H+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

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g 8. A hydrogen atom is initially a rest and in the ground state. A proton with kinetic energy 1000 eV scatters off the atom. In
aleksley [76]

Answer:

Explanation:

Ionization energy of hydrogen atom is 13.6 eV . This energy will be provided by energetic proton , the kinetic energy of which is 1000 eV.  The kinetic energy of ionized electron is 15.2 eV . Kinetic energy of proton produced from from the ionization of hydrogen or the nucleus of the hydrogen atom is 4.3 eV . All these energy must have come from kinetic energy of initial proton.

So kinetic energy of projectile proton after collision

= 1000 - ( 13.6 + 15.2 + 4.3 ) eV.

= 966.9 eV .

8 0
3 years ago
What are the names of the positive and negative electrodes of an electrolytic cell?
xz_007 [3.2K]

Answer:

the positive electrode is an anode, and the negative electrode is a cathode.

Explanation:

6 0
2 years ago
What were the results on Rutherford's goldfoil experiment
diamong [38]

Answer:

Explanation:

The relative massive alpha particles could go through the gold foil without being deviated of their trajectory or only small deviations due to the uniformity distribution positive charge of the protons.

4 0
3 years ago
A student determines that she used 0.0665 mol of sodium hydroxide (NaOH) to completely titrate 25.00 mL of sulfuric acid solutio
AysviL [449]

To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.

Let us determine the reaction for the titration below:

2NaOH +2H2SO4 = Na2SO4 +2H2O

 

So,

0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution 

= 2.62 M H2SO4


The answer is the fourth option: 

<span>2.62 M</span>
7 0
3 years ago
Read 2 more answers
How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

8 0
3 years ago
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