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3 years ago

A student performed the following reaction: 2 A + B2 --> 2AB

Physics

1 answer:

raketka [301]3 years ago

6 0

Answer:

116g

Explanation:

Given parameters:

Mass of A = 88g

Mass of B₂ = 28g

Unknown:

Mass of AB produced = ?

Solution:

The reaction expression is given as:

2A + B₂ → 2AB

According to the law of conservation of matter "matter is neither created nor destroyed but changed from one form to another".

In retrospect;

Mass of A + Mass of B₂ = Mass of AB

Now insert the parameters and solve;

Mass of AB = 88g + 28g = 116g

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If a 1-megaton nuclear weapon is exploded

Elenna [48]

Answer:

1,780,000 N

Explanation:

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Force = pressure × area

F = 20,260 Pa × (3.89 m × 22.6 m)

F = 1,780,000 N

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2 years ago

List several examples of applied force, normal force, and friction that you’ve observed in your life.

grandymaker [24]

applied forces would be push for example.

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3 years ago

Fifteen grams of substance X at 95 degrees Celsius is mixed with 45 grams of substance Z at 85 degrees Celsius in a container wh

Viefleur [7K]

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H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.

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3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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3 years ago

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valina [46]

Answer:

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3 years ago