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zheka24 [161]
2 years ago
6

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

al velocity of the car is +22.0 meters/second. After the collision the velocity of the car is -11.0 meters/second. What is the velocity of the truck after this elastic collision?
A.+13 meters/second
B.+20 meters/second
C.+15.7 meters/second
D.+7.5 meters/second
E.+9.8 meters/second
Chemistry
2 answers:
LenaWriter [7]2 years ago
7 0

Answer:

c

Explanation:

FOR PLATO

d1i1m1o1n [39]2 years ago
6 0
The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision

Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
For car:mass=2.3×10^3, velocity=-11
For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
24200 = -12100 + 2.3×10^3V
2.3×10^3V = 36300
V = 15.7 m/s
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Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

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According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

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3 0
2 years ago
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546

K

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=

T

2

=

273

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Final Vloume

=

V

2

=

?

?

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Since the pressure is constant and the question is asking about temperature and volume, i.e,

V

1

T

1

=

V

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T

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⇒

V

2

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T

2

T

1

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546

=

60

2

=

30

l

i

t

e

r

⇒

V

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