Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)
Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)
Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2
Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g
1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au
Volume = Area (of the face) * Length
We want Length:
Length = Volume/Area
L = (518.5 cm^3/0.606 cm^2)
L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)
Answer:
i wont le me down load it
Explanation:
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
Answer:
The acceleration of the car is 9,19 m/s2
Explanation:
We use the formula: F=m x a---> a=F/m
a=21,6N/ 2,35kg 1N is 1kgxm/s2
a=21,6 kg x m/s2 x 2,35 kg
a=9,191489362 m/s2
