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daser333 [38]
2 years ago
10

A compound is found to have a molar mass of 598 g/mol. if 35.8 mg of the compound is dissolved in enough water to make 175 ml of

solution at 25°c, what is the osmotic pressure of the resulting solution?
Chemistry
1 answer:
TiliK225 [7]2 years ago
7 0
The computation for molarity is: 
(x) (0.175 L) = 0.0358 g / 598 g/mol 
x = 0.000342093 M 
Whereas the osmotic pressure calculation: 
pi = iMRT 
pi = (1) (0.000342093 mol/L) (0.08206 L atm / mol K) (298 K) 
pi = 0.0083655 atm 
Converting the answer to torr, will give us:
0.0083655 atm times (760 torr/atm) = 6.35778 torr 
which rounds off to 6.36 torr
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7 0
2 years ago
Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 669.1-g sam
Vedmedyk [2.9K]

Answer:

2.4 ×10^24 molecules of the herbicide.

Explanation:

We must first obtain the molar mass of the compound as follows;

C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1

We know that one mole of a compound contains the Avogadro's number of molecules.

Hence;

169 g of the herbicide contains 6.02×10^23 molecules

Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.

7 0
2 years ago
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Answer:

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6 0
2 years ago
A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav
dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of _{19}^{40}\textrm{K} radioisotope follows:

Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles  

Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles  

_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e

By the stoichiometry of above reaction,

1 mole of _{40}^{91}\textrm{Zr} is produced by 1 mole _{41}^{91}\textrm{Nb}

So, 0.17 moles of _{40}^{91}\textrm{Zr} will be produced by = \frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}

Amount of _{82}^{212}\textrm{K}

decomposed will be = 0.17 moles

Initial amount of _{40}^{91}\textrm{Nb}  will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = 0.024

a_o = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

0.024=\frac{0.194}{2^n}

n=3

Therefore, 3 half lives have passed.

3 0
2 years ago
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A certain element has a melting point over 700 ∘C and a density less than 2.00 g/cm3. What is one possible identity for this ele
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<span>  The element is Beryllium</span>
6 0
2 years ago
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