Answer:
1. [OH⁻] = 0.30 M ; 2. [OH⁻] = 1.54x10⁻⁶M ; 3. [OH⁻] = 1.32x10⁻¹³M
Explanation:
Remember the rule:
pH + pOH = 14
pOH = 14 - pH
10*⁻pOH (you have to elevate 10, to -pOH)
10*⁻pOH = [OH⁻]
1. 14 - 13.48 = 0.52
10⁻⁰°⁵² = 0.30
2. 14 - 8.19 = 5.81
10⁻⁵°⁸¹ = 1.54x10⁻⁶
3. 14 - 2.12 = 12.88
10⁻¹²°⁸⁸ = 1.32x10⁻¹³
Answer:
Ni(NO3)2 + 2NaOH → Ni(OH)2 + 2NaNO3
NET IONIC EQUATION :
Ni +2(aq) + 2(NO3) -1(aq) + 2Na +1(aq) + 2OH -1(aq)
→ Ni(OH)2 (s) + 2Na +1(aq) + 2(NO3) -1(aq)
sorry there is no space to write the reaction in one line...
Depending on what you are doing, virtual, or in school, you can contact your school and you ask them, or you can have your teacher, based on the subject, tutor you.
<3
I think it's Letter c.13 if I'm not mistaken
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
