Answer:
c = 0.07 j/g.k
Explanation:
Given data:
Mass of sample = 35 g
Heat absorbed = 48 j
Initial temperature = 293 K
Final temperature = 313 K
Specific heat of substance = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
ΔT = 313 k - 293 K
ΔT = 20 k
Now we will put the values in formula.
48 j = 35 g × c× 20 k
48 j = 700 g.k ×c
c = 48 j/700 g.k
c = 0.07 j/g.k
This lesson is the first in a three-part series that addresses a concept that is central to the understanding of the water cycle—that water is able to take many forms but is still water. This series of lessons is designed to prepare students to understand that most substances may exist as solids, liquids, or gases depending on the temperature, pressure, and nature of that substance. This knowledge is critical to understanding that water in our world is constantly cycling as a solid, liquid, or gas.
In these lessons, students will observe, measure, and describe water as it changes state. It is important to note that students at this level "...should become familiar with the freezing of water and melting of ice (with no change in weight), the disappearance of wetness into the air, and the appearance of water on cold surfaces. Evaporation and condensation will mean nothing different from disappearance and appearance, perhaps for several years, until students begin to understand that the evaporated water is still present in the form of invisibly small molecules." (Benchmarks for Science Literacy<span>, </span>pp. 66-67.)
In this lesson, students explore how water can change from a solid to a liquid and then back again.
<span>In </span>Water 2: Disappearing Water, students will focus on the concept that water can go back and forth from one form to another and the amount of water will remain the same.
Water 3: Melting and Freezing<span> allows students to investigate what happens to the amount of different substances as they change from a solid to a liquid or a liquid to a solid.</span>
Answer:
Li and H
Explanation:
2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced
<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:
3. Equal numbers of protons and neutrons
Explanation:
The nucleus becomes unstable if the ratio of protons to neutrons is less than 1:1 or more than 1:1.5.
The most stable nucleus has a neutron proton ratio of 1:1 which means that they can not release a neutron or a proton to decay.
Nucleus 3 is therefore the most stable.
