The balanced nuclear equations for the following:(a) β⁻ decay of silicon-32 is (27,14)Si -> (0,-1)beta + (27,15)P
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What is balanced nuclear equation?</h3>
A nuclear reaction is generally expressed by a nuclear equation, which has the general form, where T is the target nucleus, B is the bombarding particle, R is the residual product nucleus, and E is the ejected particle, and Ai and Zi (where I = 1, 2, 3, 4) are the mass number and atomic number, respectively. Finding a well balanced equation is critical for understanding nuclear reactions. Balanced nuclear equations provide excellent information about the energy released in nuclear reactions. Balancing the nuclear equation requires equating the total atomic number as well as the total mass number before and after the reaction using the rules of atomic number and mass number conservation in a nuclear reaction.
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Answer:6 atoms
Explanation:Each NaOH has one Na and one O and one H. Therefore, 2 NaOH has 6 atoms.
The solution changed color because the substances are not neutral.
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pH</h3>
Chemical substances have different concentrations of the hydrogen cation, called PH.
The higher the pH, the more basic the substance, and the lower the more acidic.
Bromothymol blue is a pH indicator that changes its color according to the pH of the substance, yellow for acid, blue for basic and green for neutral.
In the case of the reactions in question, we have the release of CO2 (acid) in combustion and in cellular respiration, changing the color of bromothymol blue to yellow.
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
