The car's velocity after 5s : 10 m/s
<h3>Further explanation</h3>
Given
velocity=v=72 km/h=20 m/s
time=t = 5 s
acceleration=a = -2 m/s²
Required
velocity after 5s
Solution
Straight motion changes with constant acceleration
vf=final velocity
vi = initial velocity
Input the value :
The car is decelerating (acceleration is negative) so that its speed decreases
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
Answer:
The answer to your question is M = 36.49 g
Explanation:
Data
mass = 8.21 g
volume = 4.8064 L
Temperature = 200°C
Pressure = 1.816 atm
M = ?
Process
1.- Convert temperature to °K
°K = 273 + 200
°K = 473
2.- Calculate the number of moles
n = (PV)/RT
n = (1.816)(4.8064)/(0.082)(473)
n = 0.225
3.- Calculate the molar mass
M --------------- 1 mol
8.21 g ---------- 0.225 moles
M = (1 x 8.21)/0.225
M = 36.49 g
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