Question 4 (1 point)

a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the ha

Nana76 [90]

Answer:

Approximately $4.5\; \text{hours}$ .

Explanation:

Calculate the ratio between the mass of this sample after $22.5\; \text{hours}$ and the initial mass:

$\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125$ .

Let $n$ denote the number of half-lives in that $22.5\; \text{hours}$ (where $n\!$ might not necessarily be an integer.) The mass of the sample is supposed to become $(1/2)$ the previous quantity after each half-life. Therefore, if the initial mass of the sample is $1\; \rm g$ (for example,) the mass of the sample after $\! n$ half-lives would be ${(1/2)}^{n}\; \rm g$ . Regardless of the initial mass, the ratio between the mass of the sample after $n\!\!$ half-lives and the initial mass should be ${(1/2)}^{n}$ .

For this question:

${(1/2)}^{n}} = 0.03125$ .

Take the natural logarithm of both sides of this equation to solve for $n$ :

$\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)$ .

$n\, [\ln(1/2)] = \ln (0.03125)$ .

$\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5$ .

In other words, there are $5$ half-lives of this sample in $22.5\; \text{hours}$ . If the length of each half-life is constant, that length should be $(1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}$ .

4 0

3 years ago

Pitchblende is ore of _____.

Valentin [98]

Its an ore of uraninte i think.

6 0

4 years ago

Read 2 more answers

Determine the empirical and molecular formula for a compound containing 26.1

kobusy [5.1K]

determing that the compoumd formed is in 100%

therefore, 26.1 +4.3+69.6 = 100%

hence no need of adding O molecule

finding the numbr of moles of each element

no. of moles of C= given mass÷molar mass

=26.1÷12

=2.1 mol

no. of moles of H = given mass÷molar mass

=4.3÷1

=4.3 mol

no of moles of O =given mass÷molar mass

=69.6÷16

=4.3

the simplest whole nber of the element in the unknown compoumd is 1:2:2 of C:H:O

HENCE, the empirical formula is CH2O2

empirical formula mass of CH2O2 IS

=(12)+(2×1)+(16×2)

12+2+32

46 G/MOL

N= molar mass of the compoumd ÷ emporical mass of the compoumd

N= 138÷46

N=3

molecular formula =N×empirical formula

=3 ×CH2O2

C3H6O6 is the molecular formula

(tartonic acid)

hope it helped u!

5 0

3 years ago

the volume of a rectangular solid is the product of its length, width, and height. Calculate the volume of a rectangular solid w

nalin [4]

12.3*3.8=46.74cm2
46.74*11.4=532.836cm3

5 0

3 years ago

Read 2 more answers

Draw Au3N as a picture of atoms

Shkiper50 [21]

The picture of Au₃N is attached below.

The first part of the picture shows the formation of Au and N ions.

Formation of Au⁺¹ :

As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.

Formation of N⁻³ :

As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)

Formation of Au₃N:

Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.

7 0

3 years ago