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Furkat [3]
3 years ago
9

I think I'm typing it into my calculator wrong. I will give brainliest to whoever gets it right.

Chemistry
1 answer:
Pepsi [2]3 years ago
3 0

Answer:

36.7 mg

Explanation:

The following data were obtained from the question.

Original amount (A₀) = 65.1 mg

Rate constant (K) = 2.47×10¯² years¯¹

Time (t) = 23.2 years

Amount of substance remaining (A) =?

Thus, we can obtain the amount of substance remaining after 23.2 years as follow

ln A = lnA₀ – Kt

lnA = ln(65.1) – (2.47×10¯² × 23.2)

lnA = 4.1759 – 0.57304

lnA = 3.60286

Take the inverse of ln

A = e^3.60286

A = 36.7 mg

Therefore, the amount remaining after 23.2 years is 36.7 mg.

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Determine the volume, in liters, occupied by 0.015 molecules of oxygen at STp?
Arlecino [84]

5.58 X 10^{-25} Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.

Explanation:

Data given:

molecules of oxygen = 0.015

number of moles of oxygen =?

temperature at STP = 273  K

Pressure at STP = 1 atm

volume = ?

R (gas constant) = 0.08201 L atm/mole K

to convert molecules to moles,

number of moles = \frac{molecules}{Avagadro's number}

number of moles = 2.49 x 10^{-26}

Applying the ideal gas law since the oxygen is at STP,

PV = nRT

rearranging the equation:

V = \frac{nRT}{P}

putting the values in the rearranged equation:

V = \frac{2.49 X10^{-26}   X 0.08201 X 273}{1}

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5 0
4 years ago
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iren2701 [21]

Answer:

FALSE

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7 0
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Answer:

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Na has an atomic mass of 23.0g and O has an atomic mass of 16.0g. How many grams of Na are needed to completely react with 40.0g
dimulka [17.4K]
The grams of Na that are needed to complete  to react with 40..0 g of O2 is calculated as  below

find the moles of O2 used = mass/molar mass

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write the reacting equation
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=  5 moles x 23 g /mol= 115  moles
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3 years ago
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