Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
d. When aluminum-28 undergoes beta decay, silicon-28 is produced.
Explanation:
When the nuclei of aluminium-28 decays, it produces silicon- 28:
Aluminium ²⁸₁₃Al
Silicon 28 ²⁸₁₄Si
beta particle ⁰₋₁
²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁
This way, the mass and atomic number are conserved.
Conservation of mass number:
28 = 28 + 0, 28 = 28
13 = 14 -1 , 13 = 13
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Answer:
58.61 grams
Explanation:
Taking The molecular weight of NaCl = 58.44 grams/mole
<u>Determine how many grams of NaCl to prepare the bath solution </u>
first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution
= ( 6 * 170 ) / 1000
= 1020 / 1000 = 1.020 moles
next
determine how many grams of NaCl
= moles of NaCl * molar mass of NaCl
= 1.020 * 58.44
= 58.61 grams
Answer is: <span>unbalanced electronegativity of the hydrogens and oxygens as they share electrons.
Oxygen has greater electronegativity than hydrogen, because of that oxygen is partially negative and hydrogen is partially positive.
</span>Electronegativity<span> is a </span>chemical property<span> that describes the tendency of an </span>atom<span> to attract a shared pair of </span>electrons<span> towards itself.</span>
Answer:
Concentration for the solution is 0.153 mol/L
Explanation:
Formula for the osmotic pressure is π = M . R . T . i
where M is molarity (concentration), R the universal constant for gases and T is Absolute T° (T°C + 273)
π = Osmotic pressure.
Let's replace the data given:
3.9 atm = M . 0.082L.atm/mol.K . 310K
3.9 atm / 0.082 mol.K/L.atm . 310K = 0.153 mol/L (M)
i = Van't Hoff factor (ions from the solute dissolved in solution)
In this case, we assumed no ion pairing, so i = 1