Answer:
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.
What is the specific heat of the substance?
If it is one of the substances found in Table 8.1.1, what is its likely identity?
Answer:
Lead shows the greatest temperature change upon absorbing 100.0 J of heat.
Explanation:

Q = Energy gained or lost by the substance
m = mass of the substance
c = specific heat of the substance
ΔT = change in temperature
1) 10.0 g of copper
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the copper = c = 0.385 J/g°C


2) 10.0 g of aluminium
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the aluminium= c = 0.903 J/g°C


3) 10.0 g of ethanol
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the ethanol= c = 2.42 J/g°C


4) 10.0 g of water
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the water = c = 4.18J/g°C


5) 10.0 g of lead
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the lead= c = 0.128 J/g°C


Lead shows the greatest temperature change upon absorbing 100.0 J of heat.
Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:

<u>3. Solution</u>
Solve, substitute and compute:


Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.

<u>3. Solution</u>
This time, solve for V₂:

Substitute and compute:

You must round to 3 significant figures:

Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:

The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:

You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg

<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:

Answer:- The hydroxide ion concentration of the solution is
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
= 0.00000195 or 
So, the hydroxide ion concentration of the solution is
.