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3 years ago

14

What are Newton's three laws of motion? Are they correct?

1 answer:

muminat3 years ago

8 0

Answer:

Explanation:

Of course Newton's three laws of motion are correct, because they were verified several hundred of years ago and they continue working today, for such systems. Science is accumulative. What modern physics has done is to constraint the range of validity of those laws.

Newtons first law of motion states that everybody continues in it rest or of uniform motion in a straight line unless an external force is applied on it.

Newtons second law of motion states that acceleration produced on a body by a force is directly proportional to the applied force and inversely proportional to the mass of body.

Newtons third law of motion states that to every action there is equal and opposite reaction.

Hope it will be helpful :)

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If half of the moon is always illuminated why does its appearance from earth change?

irina [24]

Answer:

The different angles the moon has from sunlight to the earth

Explanation:

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2 years ago

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Which radioisotopes have the same decay mode and have half-lives greater than

AveGali [126]

The correct answer is (3)

I-131 and P-32

The explanation:

according to attached table:

- we can see that the half life of p 32 is 14.28d (more than one hour)

- and the half life of I-131 is 8.021 d (more than one hour)

and They both have β- decay mode and with half-lives greater than hour.

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3 years ago

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Elanso [62]

Answer:

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2 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

3 years ago

How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂

Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

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2 years ago