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3 years ago

If you prepared a 3.25 M solution of sucrose (molar mass 342 g/mole) ,

Chemistry

2 answers:

Dafna11 [192]3 years ago

7 0

Answer:

a. 0.8125 moles of sucrose

b. 277.8 g of sucrose

Explanation:

Consider these relation's value:

Molarity = Mol / Volume(L)

Mol = Molarity . Volume(L)

Volume(L) = Mol / Molarity

So, Molarity = 3.25M

Volume(L) = 0.25L

Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol

Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose

sveticcg [70]3 years ago

4 0

Answer:

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

Explanation:

Step 1: Data given

Molarity of a sucose solution = 3.25 M

Molar mass of sucrose = 342 g/mol

Step 2: Calculate moles sucrose

Moles sucrose = molarity sucrose solution * volume solution

Moles sucrose = 3.25 M * 0.25 L

Moles sucrose = 0.8125 moles sucrose

Step 3: Calculate mass of sucrose

Mass sucrose = moles sucrose * molar mass sucrose

Mass sucrose = 0.8125 moles * 342 g/mol

Mass sucrose = 277.9 grams

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

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natta225 [31]

Answer:

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Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

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$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

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1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

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