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3 years ago

After two half-lives, a sample contains 10 grams of parent isotopes. How many grams of the parent isotope were present to start?

Chemistry

2 answers:

Lunna [17]3 years ago

8 0

T=2T
m=10 g

m=m₀2^(-t/T)

m₀=m/{2^(-2T/T)=m/2⁻²

m₀=10/2⁻²=40 g

40 grams

almond37 [142]3 years ago

8 0

Theirs a total of 39.9999G at the start

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3 years ago

Calculate the ph of a solution whose(oh^-) is 4.583*10^-5 mole dm cube (pkw=14)

Sergeu [11.5K]

pH of solution = 9.661

<h3>Further explanation </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

pOH=-log[OH⁻]

$\tt pOH=-log[4.583\times 10^{-5}]\\\\pOH=5-log~4.583=4.339$

pH+pOH=pKw

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3 years ago

what is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants

Rama09 [41]

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3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

580 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

fenix001 [56]

Answer:

81.04°C

Explanation:

Heat loss by water = Heat gained by Aluminum

Heat loss by water;

H = MCΔT

ΔT = 100 - T2

M = 580g

c = 4.2

H = 580 * 4.2 (100 - T2)

H = 243600 - 2436T2

Heat ganed by Aluminium

H = MCΔT

ΔT = T2 - 24

M = 900g

c = 0.9

H = 900 * 0.9 (T2 - 24)

H = 810 T2 - 19440

243600 - 2436T2 = 810 T2 - 19440

243600 + 19440 = 810 T2 + 2436T2

263040 = 3246 T2

T2 = 81.04°C

Assumption;

Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.

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4 years ago