Answer (there are two other names for it):
- Golgi body
- Golgi apparatus
I hope this helps!
Answer:
D. O – H
Explanation:
- The electronegativity of the elements in the question is:
<em>H (2.20), S (2.58), Cl (3.16), N (3.04), O (3.44), and C (2.55).</em>
- Since, the H atom has the lowest electronegativity (2.20) and O atom has the highest electronegativity (3.44).
- <em>So, O - H pair of atoms has the highest electronegativity difference.</em>
Answer:
mechanical energy to electrical energy to light energy
Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Answer:
0.0457M is molarity of nitrate anion in the solution.
Explanation:
Molarity is defined as the ratio between moles of solute (In this case, nitrate ion) and liters of solution (The total volume of the solution is 300.0mL = 0.300L).
Thus, we need to convert mass of Lead(II) nitrate to moles using its molar mass (Molar mass Pb(NO₃)₂: 331.2 g/mol). Moles of Pb(NO₃)₂ = 1/2 Moles of NO₃⁻:
<em>Moles Pb(NO₃)₂ and moles of NO₃⁻:</em>
2.27g * (1mol / 331.2g) = 0.006854 moles Pb(NO₃)₂ * (2moles NO₃⁻ / 1mol Pb(NO₃)₂) = 0.0137moles NO₃⁻
Molarity is:
0.0137 moles NO₃⁻ / 0.300L =
<h3>0.0457M is molarity of nitrate anion in the solution </h3>