Option C. The object is returning to the start at a constant speed.
<h3>
Data points of the Position vs Time graph</h3>
The following data points will be used to determine the motion of the object.
<u>Position Time</u>
12 4
10 6
2 8
0 10
From the data above, the position of the object is decreasing towards zero or start point.
Thus, the object is returning to the start at a constant speed.
Learn more about position here: brainly.com/question/2364404
#SPJ1
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer: ^_^
Explanation:
its because during day time the ball becomes heated up and the air inside it try to come out as the heat air always try to go up swelling up the ball,
so in the evening when the air cools down the ball too cools down and air inside it also cools down making the ball feel soft,
Therefore, a football feels hard on a hot day but feels softer in the evening
Hope it helped u,
pls mark as the brainliest
^_^
<span>Answer
is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643
1/s.
k</span>₂ = 0,00828
1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K =
0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
The third option because the elements mentioned in that selection are in the same group, they must have similar properties.