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ruslelena [56]
3 years ago
15

Just wanted to know what the response time is :)

Chemistry
2 answers:
vladimir1956 [14]3 years ago
4 0
Bshsibeidbeifhdbeuu hi chh in Chen but by uh hi in vcucbxbxbxbxbcbcbcbcbcbdjcjfjc
Nady [450]3 years ago
4 0
Too fast
I get my answers in just 2-3mins or atleat 30sec
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What is the formula for barium iodide?
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A(n) ____________________ is a compound that often tastes bitter. <br> a. acid <br> b. base
vovangra [49]
A base tastes bitter, while an acid tastes sour. 

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4 years ago
radioisotope actinium 225 has half life of 10 days, I begin with 16 kg of this isotope, how much will remain after 40 days?
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When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi
Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point  

K_f = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

K_f=2.07

ii) 20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

4 0
3 years ago
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