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olya-2409 [2.1K]

3 years ago

10

If one ball with a momentum of 50 kg·m/s hits a ball at rest and no other forces are acting on either ball, what's the total mom

entum of both balls after the collision?
A. 0 kg·m/s
B. 25 kg·m/s
C. 100 kg·m/s
D. 50 kg·m/s

1 answer:

vitfil [10]3 years ago

5 0

Answer:

The answer is (D) 50 kg·m/s (PF Test)

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What does the zeroth law of thermodynamics allow us to define?

11Alexandr11 [23.1K]

The Zeroth Law of Thermodynamics<span> states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. Credit: Tim Sharp</span>

5 0

3 years ago

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When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)

Firdavs [7]

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ = 0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂ + 2K₃PO₄ → Ba₃(PO₄)₂ + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

Ba(NO₃)₂ : Ba₃(PO₄)₂

3 : 1

0.182 × 10⁻³ : 1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

K₃PO₄ : Ba₃(PO₄)₂

2 : 1

2.537 × 10⁻³ : 1/2 × 2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by Ba(NO₃)₂ are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

6 0

3 years ago

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

stealth61 [152]

<u>Answer:</u> The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

<u>Explanation:</u>

We are given a function that calculates the amount of sample remaining after 't' years, which is:

$A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}$

<u>For t = 20 years</u>

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}$

$A_t(t)=288.522g$

Hence, the amount of sample left after 20 years is 288.522 g

<u>For t = 50 years</u>

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}$

$A_t(t)=144.26g$

Hence, the amount of sample left after 50 years is 144.26 g

6 0

3 years ago

Un ácido nítrico tiene una densidad de 10% ,calcular el peso de 1000 centímetros cúbicos

devlian [24]

D = M/V
D = 0.10
V = 1000 cm^3

0.10 = M / 1000
M = 100

Porque la densidad es 10%, se puede usar 0.10 en este formulario para calcular el peso. No sé lo que es la unidad para el peso pero es 100.

4 0

3 years ago

Mutations that neither benefit nor harm the organism have (blank) effect on the organisms survival

Daniel [21]

Answer:

I would say that the mutation has no effect on the organism, as it doesn't help or harm it.

hope this helps :)

Explanation:

3 0

3 years ago

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