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3 years ago

How would you prepare 1 liter of .030 m copper 2 sulfate pentahydrate?

1 answer:

5 0

Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O) = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.

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Answer:

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That would mean that 26.9 is the length.

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3 years ago

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4 years ago

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3 years ago

What is the half-life of an isotope that decays to 12.5% of its original activity in 19.8 hours?

katen-ka-za [31]

$x= x_{0} e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t
$
$and~ x_{0} ~the~original~amount.$
$when~x= \frac{12.5}{100} x_{0} }= \frac{125}{1000} x_{0}$
$\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}$
$when t=19.8 hrs,
x= x_{0} e^{19.8 k}, 
 \frac{x}{ x_{0} } = e^{19.8 k}, 
 \frac{1}{8} = e^{19.8k} ,
ln \frac{1}{8} =19.8 k,
$
$ln1-ln8=19.8k,
0-ln 2^{3} =19.8 k,
-3 ln2=19.8k
k= \frac{-3 ln2}{19.8} = \frac{- ln 2}{6.6}$ $x= x_{0} e^{ \frac{-ln2}{6.6}t } ,
 \frac{x}{ x_{0} } = e^{ \frac{- ln 2}{6.6}t } 
$ $\frac{1}{2} = e^{ \frac{- ln2}{6.6}t } = \frac{1}{ e^{ \frac{ln 2}{6.6} t} } ,
2= e^{ \frac{ln2}{6.6}t } ,
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8 0

4 years ago

An organic acid was analyzed and the following percent composition was obtained: 68.85% carbon, 4.9% hydrogen, and 26.2% oxygen.

rewona [7]

Answer:

The molecular formula of the compound is: C14H12O4

Explanation:

Step 1: find molality

ΔT = Kf x m

ΔT = Tfinal - Tinitial = 3.37 - 5.50 = -2.13

molality (m) = ΔT/Kf = -2.13/-5.12 = 0.416 m

step 2: find number of moles

m = number of moles (n)/Kg solvent which is the benzene

n = molality x Kg solvent = 0.416 molality X (10 x 10^-3 Kg) = 0.00416 mole

Step 3: find molecular mass of the organic compound

n = mass/Molecular mass = m/Mm

Mm = m/n = 1.02/0.00416 = 245.19 g/mole of organic compound molecular formula

Step 4: find number of mole of each compound

let consider it being in 100g where each % will correspond to the mass of the element in 100g of the compound.

C = mass/molar mass = 68.85/12.01 = 5.7327 mole

H = mass/molar mass = 4.9/1.01 = 4.8515 mole

O = mass/molar mass = 26.2/16 = 1.6375 mole

Step 5: divide by smallest number of mole

C = 5.7327/1.6375 = 3.5 = 7/2

H = 4.8515/1.6375 = 3

O = 1.6375/1.6375 = 1

C7/2H3O1

multiply by two to remove the fraction: C7H6O2 in which the empirical molecular mass = 122.13 g/mole

to find the factor for the molecular formula x = Mm of molecular/Mm of empirical = 245.19/122.13 = 2

Multiply the empirical by 2 = 2 x (C7H6O2) = C14H12O4

8 0

4 years ago