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Salsk061 [2.6K]
3 years ago
9

A 21.0 kg child is running at a velocity of 2.50 m/s. The child is carrying a 2.25 kg gallon of milk. What is the momentum of th

e child and the milk together?
Chemistry
1 answer:
KIM [24]3 years ago
7 0

Answer: The momentum of the child and milk together is 58.125 kg.m/s

Explanation:

Momentum is defined as the product of object's mass and velocity.

Mathematically,

p=mv

where, p = momentum

m = mass of the object

v = velocity of the object

In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:

p=(m_{child}+m_{milk})v     ....(1)

Given:

m_{child}=21.0kg\\m_{milk}=2.25kg\\v=2.5m/s\\p=?kg.m/s

Putting values in equation 1, we get:

p=(21+2.25)kg\times 2.5m/s=58.125kg.m/s

Hence, the momentum of the child and milk together is 58.125 kg.m/s

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How many grams are needed to prepare 750. mL of a 0.35 M solution of NaOH?
nikdorinn [45]

Answer:

10.5grams

Explanation:

Molarity = number of moles (n)/ volume (V)

According to this question;

Volume = 750 mL = 750/1000 = 0.75L

Molarity = 0.35M

number of moles (n) = molarity × volume

n = 0.35 × 0.75

n = 0.2625mol

Using mole = mass/molar mass

Molar Mass of NaOH = 23 + 16 + 1

= 40g/mol

mole = mass/molar mass

0.2625 = mass/40

mass = 10.5grams

10.5 grams are needed to prepare 0.75L of a 0.35 M solution of NaOH.

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3 years ago
What kind of data does regional climate mainly collect and concentrate on?
hram777 [196]

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Climate, atmosphere, and land

Explanation:

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8 0
2 years ago
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Musya8 [376]
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The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

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on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

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