Answer:
umm.. B. a base that generates a lot of hydroxide ions in water.
273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Answer:
Explanation:
HCl + NaOH ⟶ NaCl + H₂O
There are two energy flows in this reaction.
Heat of reaction + heat to warm water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m(HCl) = 50 g
m(NaOH) = 50 g
T₁ = 22 °C
T₂ = 28.87 °C
C = 4.18 J·°C⁻¹g⁻¹
Calculations:
m = 50 + 50 = 100 g
ΔT = 28.87 – 22 = 6.9 °C
q₂ = 100 × 4.18 × 6.9 = 2900 J
q₁ + 2900 = 0
q₁ = -2900 J
The negative sign tells us that the reaction produced heat.
The reaction produced 
.
2.10 x 10^-10 M. Ans
pH + pOH = 14
Where, pOH is the power of hydroxide ion concentration and pH is the power of concetration of the H+ ion.
Now, pOH = 14 - 4.32
= 9.68
Now, the concentration of [H+] is 10-7 M, then pH is 7 and for [OH-] = 10-7 M, the pOH is also 7.
Now, pOH = -log[OH-]
[OH-] = 10^- pOH
= 10^-9.68
= 2.10 x 10^-10 M
Answer:
The answer to your question is: 1, 2, 1, 2
Explanation:
1 Fe(s) + 2 Na⁺(aq) → 1 Fe²⁺(aq) + 2 Na(s)
Fe⁰ - 2e⁻ ⇒ Fe⁺² Oxidases
Na⁺ + 1 e⁻ ⇒ Na⁰ Reduces
1 x ( 1 Fe⁰ ⇒ 1 Fe⁺²) Interchange number of
2 x ( 2Na⁺ ⇒ 2 Na⁰ ) electrons
