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Virty [35]
8 months ago
13

Barium-131 is used in the detection of bone tumors. The half-life of barium-131 is approximately 12 days. How long will it take

for the radiation level of barium-131 to drop to 1/4 of its original level?.
Chemistry
1 answer:
Evgesh-ka [11]8 months ago
5 0

Barium-131's radiation level won't reach 1/4 of its initial level for 24 hours.

ln[A] t = -kt + ln[A] 0 is the integrated rate rule for the first-order reaction A's products.

A straight line is produced when the natural log of [A] is plotted as a function of time since this equation has the form y = mx + b.

How is the length of a half-life determined?

The amount of time needed for the reactant concentration to drop to half its initial value is known as the half-life of a reaction. A first-order reaction's half-life is a constant that is correlated with its rate constant:

t 1/2 = 0.693/k.

To know more about rate constant, visit:

brainly.com/question/20305871

#SPJ4

You might be interested in
If an atom has 30 electrons and 35 neutrons, what is its approximate atomic mass? what is the name of this element?
Murljashka [212]
30 electrons = atomic number
Look at your periodic table. You will see zinc has a atomic number (the top number) of 30.

When you add 30 and 35 you get 65. The is the atoms ATOMIC MASS. So looking at zinc on the periodic table, look at its bottom number, which is its atomic mass. It will be 65.

This means the name of the element is ZINC.

Hope this helped!
6 0
2 years ago
In a previous problem, the empirical formula of a compound of phosphorus and oxygen was found to be p2o5. experimentation shows
agasfer [191]
The empirical formula for the compound is - P₂O₅
the empirical formula is the simplest ratio of whole numbers of components in a compound.
molecular formula is the actual ratio of components in a compound.
we have to first find the number of empirical units in the molecular formula 
molecular mass - 283.89 g/mol
mass of empirical formula - 283.8 g
number of empirical units - 283.89 g/mol / 283.8 g
number of empirical units - 1.000 
therefore empirical formula = molecular formula 
molecular formula - P₂O₅
3 0
3 years ago
Calling all chemists...please help me!  Ive attached a screen shot of the problem
Olin [163]
This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.

Hope this helps!
6 0
2 years ago
Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

7 0
1 year ago
Read 2 more answers
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
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