Answer:
Percent error = 1.5%
Explanation:
Given data:
Measured value of density of graphite = 2.3 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = [Measured value - Actual value / actual value] × 100
Actual/accepted value of density of graphite = 2.266 g/cm³
Now we will put the values:
Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100
Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100
Percent error = 0.015 × 100
Percent error = 1.5%
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Answer:
Scientists use the term magma for molten rock that is underground and lava for molten rock that breaks through the Earth's surface.
Answer:
B. Thicker layer of blubber
Explanation:
For the leopard seals to survives this harsh environment, it must have a thicker layer of blubber
Blubber is a thick layer of fat. It is called the adipose tissues and found in most marine organisms.
- This layer helps in insulating the body against heat loss.
- By so doing, the animal is able to conserve internal heat.
- They have low thermal conductivity and do not easily lose heat or gain heat as such.
