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Gelneren [198K]
3 years ago
8

The kinetic energy of a ball with a mass of 0.5kg and a velocity of 10 m/s is ?

Chemistry
1 answer:
KATRIN_1 [288]3 years ago
3 0

Answer:

K.E = 25 J

Explanation:

Given data:

Mass of ball = 0.5 g

Velocity of ball = 10 m/s

Kinetic energy = ?

Solution:

Formula:

K.E = 1/2 mv²

m = mass

v = velocity

Now we will put the values in formula.

K.E = 1/2 mv²

K.E = 1/2× 0.5kg × (10m/s)²

K.E = 1/2 ×0.5kg × 100m²/s²

K.E = 25 Kg.m²/s²

Kg.m²/s² = J

K.E = 25 J

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8 0
3 years ago
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
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