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viktelen [127]
3 years ago
8

What is the molarity of a solution prepared by diluting 43.72 ml of 1.005 M aqueous K2CR2O7 to 500 ml

Chemistry
1 answer:
UNO [17]3 years ago
3 0

Answer:

43.72

Explanation:

that is the answer hope u liked it and I did this already along time ago

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What elements can join wit other elements o form a covalent bond
Bas_tet [7]

Answer:

Non metals join to form covalent bond.

Explanation:

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

Both atoms bonded through covalent bond.

In Cl₂ both chlorine atoms are bonded through the covalent bond.

8 0
3 years ago
The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye
ryzh [129]

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

ln \frac{N}{N_0}  = - k*t ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

T _\frac{1}{2} = \frac{ln2}{k} ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

Multiply both side by k

28.8 yrs * k = \frac{ln 2 }{k} * k

Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

3 0
3 years ago
Read 2 more answers
Which of the following is the correct isotope notation for the element above?
ozzi
7 becuse it splits in half
6 0
3 years ago
Read 2 more answers
Lysine is an amino acid that is an essential part of nutrition but which is not synthesized by the human body. What is the molar
nata0808 [166]

Answer:

The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.

Explanation:

The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.

The parameters of this equation are:

P = 1.918 atm

V = 750.0mL = 0.75L

n = ?

R = 0.0821

T = 25 degree celcius = 25 + 273 = 298 degree kelvin.

From this formular, n = (PV)/(RT)

n = (1.918 X 0.75)/(0.0821 X 298 )

n = 0.0588

n, no of mole = mass/molar mass

0.0588 = 8.6/MM

MM = 8.6/0.0588

MM = 146.25g/mole.

4 0
3 years ago
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A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the
Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  


5 0
2 years ago
Read 2 more answers
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