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2 years ago

KI + Cl2 ---> KCl + I2 Balance the single replacement chemical reaction.

Physics

1 answer:

snow_lady [41]2 years ago

5 0

Answer:

2KI + Cl₂ → 2KCl + I₂

Explanation:

The reaction equation is given as:

KI + Cl₂ → KCl + I₂

The problem at hand is to balance this chemical reaction. To solve this problem we use a mathematical approach;

aKI + bCl₂ → cKCl + dI₂

Conserving K : a = c

I : a = 2d

Cl : 2b = c

Now let a = 1, c = 1 , d = $\frac{1}{2}$ , b = $\frac{1}{2}$ , ;

Multiply through by 2;

a = 2, b = 1 , c = 2, d = 1

2KI + Cl₂ → 2KCl + I₂

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A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an a

djverab [1.8K]

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

F = K x 8Q x 2Q / r² where k is a constant .

F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

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2 years ago

The ocean's level is currently rising at about

chubhunter [2.5K]

Answer:

The sea level will be 5m higher in 1667 y (years)

Explanation:

From the question, the rate at which the ocean's level is currently rising is about 3mm per year.

First, we will convert mm (millimeter) to m (meter)

1 mm = 0.001 m

Then,

3 mm = 3 × 0.001 m

= 0.003m

That is, the rate at which the ocean's level is currently rising is about 0.003m per year.

Now, to determine how long it will take for the ocean's level be 5 m higher than now at the given rate,

If the ocean rises 0.003 m in 1 year, then

the ocean will rise 5 m in x years

x = (5 m × 1 year) / 0.003 m

x = 5 / 0.003

x = 1666.67 years

x ≅ 1667 years

Hence, the sea level will be 5m higher in 1667 y (years)

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3 years ago

What is a reasonable measurement for the distance to Neptune?

IRINA_888 [86]

Answer:

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3 years ago

Read 2 more answers

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

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3 years ago

how much energy is required to move a 4.00-microcoulomb charge through a potential difference of 36.0 volts?

musickatia [10]

Answer:

All you gotta do is search it up. ask the same question you asked on here, type it into google, it'll teach you a lot more than your actual teachers.

Explanation:

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3 years ago