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anastassius [24]

3 years ago

9

How much faster will neon effuse than krypton, given that the molar mass of krypton is 83.8 grams and that of neon is 20.18 gram

s?

1 answer:

JulijaS [17]3 years ago

4 0

Use graham's law of diffusion

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Doctors use the radioactive isotope chromium-51 to label red blood cells in the human body. Chromium-51 gives off relatively ___

Airida [17]

Alpha particles are relatively heave and can be stopped by a sheet of paper.

7 0

3 years ago

Read 2 more answers

100 points+Brainliest for correct answer

morpeh [17]

Answer:

O Option 1

Explanation:

IF ENERGY IS RELEASED, THEN ENERGY RELEASED SHOULD BE SUBTRACTED FROM ORIGINAL.

(16.32 X 10^-19) - (5.4 X 10^-19)

10.92 X 10^-19

6 0

2 years ago

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What is 93,000,000 in scientific notation?

DanielleElmas [232]

I believe it is 9.3x10^7

5 0

3 years ago

What reaction might we use to synthesize nickel sulfate,niso4? Write equations for acid base neutralization. Include states in y

il63 [147K]

Answer:

Here's what I get

Explanation:

1. Nickel sulfate

base + acid ⟶ salt + water

NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.

Hydroxides of transition metals are insoluble; most sulfates are soluble.

$\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}$

2. Carbonate + acid

Most carbonates are insoluble.

They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

$\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)$

5 0

3 years ago

Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y

olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$A=A_o\times e^{-\lambda t}$

Where:

$\lambda$ = decay constant

$A_o$ =concentration left after time t

$t_{\frac{1}{2}}$ = Half life of the sample

Half life of Pu-239 = $t_{\frac{1}{2}}=24,000 y$ [

$\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]$

Let us say amount present of Pu-239 today = $A_o=x$

A = ?

$A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}$

$A=0.9715\times x$

$\frac{A}{A_0}=\frac{A}{x}=0.9715$

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0

3 years ago