ANSWER THESE 3 QUESTION AND I WILL GIVE OUT THE BRAINLIEST
2 answers:
<h2>
</h2>
</h2></h2></h2></h2></h2>
What is a positive acceleration ?
<h2>If the velocity of an object increases with time, its acceleration is positive.
<h2 /><h2>What is a zero acceleration ?
<h2>Zero acceleration is an acceleration, just with zero magnitude. Motion with constant velocity is just a special case of motion with uniform acceleration.
<h2>What is a negative acceleration ?
<h2>A negative acceleration means you will subtract from the current value of the velocity, and a positive acceleration means you will add to the current value of the velocity. ... And if the acceleration points in the opposite direction of the velocity, the object will be slowing down.
You might be interested in
Answer:
Limiting reagent = lead(II) acetate
Theoretical yield = 1.2704 g
% yield = 78.09 %
Explanation:
Considering:
Or,
Given :
For potassium sulfate :
Molarity = 0.120 M
Volume = 57.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 57.0×10⁻³ L
Thus, moles of potassium sulfate:
Moles of potassium sulfate = 0.00684 moles
For lead(II) acetate :
Molarity = 0.118 M
Volume = 35.5 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 35.5×10⁻³ L
Thus, moles of lead(II) acetate :
Moles of lead(II) acetate = 0.004189 moles
According to the given reaction:
1 mole of potassium sulfate react with 1 mole of lead(II) acetate
0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate
Moles of lead(II) acetate = 0.004189 moles
Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)
The formation of the product is governed by the limiting reagent. So,
1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate
0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate
Molar mass of lead(II) sulfate = 303.26 g/mol
Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g
Theoretical yield = 1.2704 g
Given experimental yield = 0.992 g
<u>% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %</u>