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GrogVix [38]
3 years ago
10

Isotope b has a half-life of 3 days. a scientist measures out 100 grams of this substance. after 6 days has passed, the scientis

t reexamines the sample.
Chemistry
1 answer:
Brilliant_brown [7]3 years ago
6 0
So half life is the time taken for a sample to decay to half its original mass, its a constant and applies to any original mass, it could be 5g or 1kg, it will take the same amount of time for the original mass to half. In this case the half life is 3 days.

After 3 days the sample will be at half its original mass, now 50g. 

Now we can treat the 50g as if its a new sample. After another 3 days (6 days in total) there will be half of 50g left, = 25g. 


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CaBr conducts electricity in the molten state but does not conduct as a solid. ionic dissolution equation.
7 0
2 years ago
The empirical formula of glucose is CH2O the molecular formula of glucose is 6 times than the empirical formula. what is the mol
jeka57 [31]

Answer:

The molecular formula of glucose is C₆H₁₂O₆

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of smallest whole number of atoms.

Molecular formula:

It gives the total number of atoms in a molecule of compound.

The molecular formula and empirical formula can be related as follow:

Molecular formula = n × empirical formula

Given data:

Empirical formula = CH₂O

Molecular formula = ?

It is stated in given problem that molecular formula is the 6 times of the empirical formula.

Molecular formula = n × empirical formula

Molecular formula = 6 × CH₂O

Molecular formula = C₆H₁₂O₆

The molecular formula of glucose is C₆H₁₂O₆.

5 0
3 years ago
For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

     2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

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