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Vaselesa [24]
4 years ago
8

The half-life of Palladium-100 is 4 days. After 12 days a sample of Palladium-100 has been reduced to a mass of 4 mg. What was t

he initial mass (in mg) of the sample?
Chemistry
1 answer:
Degger [83]4 years ago
3 0

Answer : The initial mass of the sample is, 31.9 mg

Explanation :

Half-life = 4 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{4\text{ days}}

k=0.173\text{ days}^{-1}

Now we have to calculate the initial mass of sample.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.173\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 4 mg

Now put all the given values in above equation, we get

12=\frac{2.303}{0.173}\log\frac{a}{4}

a=31.9mg

Therefore, the initial mass of the sample is, 31.9 mg

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What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{55\ g}{26.981539\ g/mol}

Moles= 2.0384\ mol

According to the reaction:-

4Al+3O_2\rightarrow 2Al_2O_3

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with \frac{3}{4} moles of oxygen gas

2.0384 moles of aluminum react with \frac{3}{4}\times 2.0384 moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

7 0
3 years ago
Comprised of protons, neutrons, and electrons.
dsp73

Answer:

The atom

Explanation:

6 0
3 years ago
sucrose is composed of carbon hydrogen and oxygen which of the following is not needed in order to determine the precent by mass
Dimas [21]

Explanation:

It is often desirable to determine the mass percent of elements in a given compound.

To determine the mass percent of elements:

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3 years ago
Chlorophyll a is one of the green pigments found in plants. Chlorophyll a has the molecular formula C55H72MgN4O5. How many atoms
yuradex [85]
We add up all the various atoms:
C: 55
H: 72
Mg: 1
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O: 5

55 + 72 + 1 + 4 + 5
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The answer is B.
3 0
3 years ago
Read 2 more answers
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
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