Answer:
16 °C
Explanation:
Step 1: Given data
- Provided heat (Q): 811.68 J
- Mass of the metal (m): 95 g
- Specific heat capacity of the metal (c): 0.534 J/g.°C
Step 2: Calculate the temperature change (ΔT) experienced by the metal
We will use the following expression.
Q = c × m × ΔT
ΔT = Q/c × m
ΔT = 811.68 J/(0.534 J/g.°C) × 95 g = 16 °C
Wind speed bc The faster the wind, the longer it blows, or the farther it can blow uninterrupted, the bigger the waves.
Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml
Answer:
406.45mL
Explanation:
The following data were obtained from the question:
V1 = 350mL
P1 = 720mmHg
P2 = 630mmHg
V2 =?
The new volume can be obtain as follows:
P1V1 = P2V2
720 x 350 = 620 x v2
Divide both side by 620
V2 = (720 x 350) /620
V2 = 406.45mL
The new volume of the gas is 406.45mL
