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2 years ago

Imagine the following scenario involving marbles as "radioactive isotopes".

Chemistry

1 answer:

sveta [45]2 years ago

8 0

Answer:

1 million years

Explanation:

The half life of a radioactive isotope refers to the time it takes for the radioactive isotope to decay to half of its original amount.

The half life of a radioactive isotope is independent of the amount of starting material. Hence, whether the amount of starting material is small or large, the half life of the substance remains the same.

Hence the half life of 2 green marbles and 500,000 green marbles is 1 million years.

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If you throw a ball straight up into the air, when is the kinetic energy the greatest?

skad [1K]

Answer: when its in the air

Explanation:

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2 years ago

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What happens to the electrons in an ionic bond

asambeis [7]

Answer:

Result of the formation of positive and negative ions.

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2 years ago

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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

2 years ago

Sodium electron configuration in longhand notation

Elden [556K]

Answer:

Na₁₁ = 1s² 2s² 2p⁶ 3s¹

Explanation:

Sodium is present in group 1.

It is alkali metal.

It has one valence electron.

The atomic number of sodium is 11.

Its atomic mass is 23 amu.

The longhand notation of electronic configuration of sodium can be written as,

Na₁₁ = 1s² 2s² 2p⁶ 3s¹

The electronic configuration in shorthand notation( noble gas) would be written as,

Na₁₁ = [Ne] 3s¹

Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.

It react with halogen and form salt. Such as sodium chloride.

2Na + Cl₂ → 2NaCl

5 0

2 years ago

In a closed system containing three different gases as shown, X2 + Y2 = 2XY

Luden [163]

X2+Y2-2XY=(X-Y)2=0
X=Y

5 0

2 years ago