A runner stood at the 100-meter mark on a track. When the timer started, the runner sprinted north to the 200-meter mark. It too
1 answer:
Answer:
4m/sec north
Explanation:
Given the runner sprinted from 100m mark to 200m mark, which means the displacement of the runner is 200-100=100m.
Now given the time taken to travel this 100 distance is 25 seconds
We know that Average velocity =
Average velocity = = 4 m/sec.
We can observe from the above formulae that the direction of the average velocity is in the direction of the displacement which is given north.
Therefore the Average velocity of the runner is 4m/sec north
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Answer:
a) 0,5 mol O₂; 1 mol NO₂
b)
Liters of NO = 22,4 L
Liters of O₂ = 11,2 L
Liters of NO₂ = 22,4 L
c) NO = 30g
O₂ = 16g
NO₂ = 46g
d) 7,41g HCl
e) 107,7 g/mol
f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL
ii. 51%
g) 32,6L
Explanation:
a) For the reaction:
2 NO + O₂ → 2 NO₂
For 1 mole of NO there are consumed:
1 mol NO × = <em>0,5 mol of O₂</em>
And produced:
1 mol NO × = <em>1 mol of NO₂</em>
b) By ideal gas law:
V = nRT/P
Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:
Liters of NO = 22,4 L
Liters of O₂ = 11,2 L
Liters of NO₂ = 22,4 L
c) The mass of each compound are:
1 mol NO× = <em>30g</em>
0,5 mol O₂× = <em>16g</em>
1 mol NO₂× = <em>46g</em>
d) Using:
n = PV / RT
Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:
0,203 moles of HCl. In grams:
0,203 mol HCl× = <em>7,41 g of HCl</em>
e) Using:
δRT/P = MW
Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)
And MW is molecular mass: <em>107,7 g/mol</em>
f) For the reaction:
2 KClO₃ → 2 KCl + 3 O₂
2,550 g of KClO₃ are:
2,550 g of KClO₃× = 0,0208 moles of KClO₃
When these moles reacts completely produce:
0,0208 moles of KClO₃× = <em>0,0312 moles of O₂</em>
In grams:
0,0312 moles of O₂ × = <em>0,999g of O₂</em>
V = nRT/P
At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; <em>V = 0,699L ≡ 699mL</em>
At 29 °C (302,15K) and 732 torr (0,963 atm)
<em>V = 0,803L ≡ 803mL</em>
ii. 182 mL ≡ 0,182L of O₂ are:
n = PV/RT
moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:
7,07x10⁻³ moles of O₂× = 0,0106 mol KClO₃. In grams:
0,0106 moles of KClO₃× = 1,300 g of KClO₃.
Thus, percent by mass of KClO₃ in the mixture is:
1,300g/2,550g ×100 = <em>51%</em>
g. Combined gas law says that:
Where:
P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) <em>V₂ = 32,6 L</em>
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I hope it helps!