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3 years ago

PLEASE HELP

Chemistry

2 answers:

MAXImum [283]3 years ago

7 0

Answer:

Alloys. And can be represented by one chemical symbol

Explanation: im just sMeRT LikE thAT yuo KnoW

n200080 [17]3 years ago

6 0

Mk i will comment so u can make them brainliest lols

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Phosphorus-32 (32P) is an isotope that is commonly used for medical and biological research. Phosphorus-32 has a half-life of 14

Pachacha [2.7K]

The activity of the sample when it was shipped from the manufacturer is 4.54 mCi

<h3>How to determine the number of half-lives that has elapsed </h3>

From the question given above, the following data were obtained:

Time (t) = 48 hours
Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
Number of half-lives (n) =?

n = t / t½

n = 48 / 342.72

n = 0.14

<h3>How to determine the activity of the sample during shipping </h3>

Number of half-lives (n) = 0.14
Original activity (N₀) = 5.0 mCi
Activity remaining (N) =?

N = N₀ / 2ⁿ

N = 5 / 2^0.14

N = 4.54 mCi

Thus, the activity of the sample during shipping is 4.54 mCi

Learn more about half life:

brainly.com/question/2674699

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2 years ago

Rsy-feiu-xad j0in only girlś

kirill [66]

Answer:

nahhhhhhhhhhhhhhhhhhhhhhh xD

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Explanation:

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3 years ago

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How many ways can you choose ten tadpoles from a pool of 20

Georgia [21]

Take half the amount

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3 years ago

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kolbaska11 [484]

Answer:

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Explanation:

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5 0

2 years ago

There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc

zhuklara [117]

Answer:

$K_{sp}$ of $Zn(CH_{3}COO)_{2}$ is 0.0513

Explanation:

Solubility equilibrium of $Zn(CH_{3}COO)_{2}$ :

$Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}$

Solubility product of $Zn(CH_{3}COO)_{2}$ ( $K_{sp}$ ) is written as- $K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}$

Where $[Zn^{2+}]$ and $[CH_{3}COO^{-}]$ represents equilibrium concentration (in molarity) of $Zn^{2+}$ and $CH_{3}COO^{-}$ respectively.

Molar mass of $Zn(CH_{3}COO)_{2}$ = 183.48 g/mol

So, solubility of $Zn(CH_{3}COO)_{2}$ = $\frac{43.0}{183.48}M$ = 0.234M

1 mol of $Zn(CH_{3}COO)_{2}$ gives 1 mol of $Zn^{2+}$ and 2 moles of $CH_{3}COO^{-}$ upon dissociation.

so, $[Zn^{2+}]$ = 0.234 M and $[CH_{3}COO^{-}]$ = $(2\times 0.234)M=0.468M$

so, $K_{sp}=(0.234)\times (0.468)^{2}=0.0513$

8 0

3 years ago