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GarryVolchara [31]
2 years ago
5

2H, +0,-> 2H,0 How many hydrogen atoms are on the left side of the equation?

Physics
1 answer:
marin [14]2 years ago
8 0

Answer:

4

Explanation:

The reaction between hydrogen and oxygen gives water (H2O)

2H,+O, -> 2H2O

Oxygen is a gas O, = O2

Balancing both sides

let , be x

2x = 4

x =  \frac{4}{2}

x= 2

Therefore, 2H2 = 4 hydrogen

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A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0
3 years ago
A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

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