Answer:
660J/kg.°C
Explanation:
Using the fact that
∆E of metal + ∆E of water = 0
∆E of the metal = Q = m*c*(T2-T1)
Where
Q = heat lost or gained = ?
m = mass of metal in kg = 0.076kg
c = specific heat of the metal = ?
T1 = 96°C
T2=31°C
∆E of the water=Q= m*c*(T2-T1)
Where
Q = 0.120*4180*6.5 = 3260J
Therefore
3,260J + Q of the metal =0
3,260J + 0.076*C*(31.0–96.0)=0
3,260J - 4.94c = 0
3,260 = 4.94c
3,260/4.94 = c
660J/kg.°C = specific heat of the metal
hope this helps:)
<h3>
Answer:</h3>
45°C
<h3>
Explanation:</h3>
We are given;
- Quantity of heat = 1.95 kJ
- Mass of lead = 500 g
- Initial temperature = 15° C
- heat capacity of lead is 130 J/kg°C
We are required to calculate the final temperature;
- We need to know that the quantity of heat is given by the formula;
Q = mass,m × specific heat,c × Change in temperature,ΔT
ΔT = final temperature - initial temperature
We can rearrange the formula to calculate the change in temperature
ΔT = Q ÷ (m×c)
Therefore;
ΔT = 1950 J ÷ (0.5 kg × 130 J/Kg°C)
= 30°C
But,
Final temperature = ΔT + Initial temperature
= 15°C + 30°C
= 45°C
Therefore, the final temperature is 45°C
Answer:
165 ml
Explanation:
We are given;
Initial volume; V_a = 55 ml
Initial molarity; M_a = 3 M
Molarity of desired solution; M_b = 0.75 M
Volume of desired solution; V_b = (55 + x) ml
Where x is the volume of water to be added.
To solve for V_b, we will use the equation ;
M_a•V_a = M_b•V_b
V_b = (M_a•V_a)/M_b
V_b = (3 × 55)/0.75
V_b = 220 mL
Thus;
(55 + x) = 220
x = 220 - 55
x = 165 mL
Answer:
False
Explanation:
Metals are materials that exist in the form of solid (state of matter). It is not true that solid solutions are only formed with metals. A solid solution is a <u>solid-state solution of one or more solutes in a solvent. </u>The solid solution can be formed with different forms. In general, solid solutions <u>formed by mixing a foreign elements Y (solute) with a perfect crystalline element X (solvent) </u>and atoms of both X and Y share the various crystal sites.