What type of reaction is C 12 H 22 O 11> 12C+11H2O

he mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reaction

valina [46]

Answer:

$MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

Explanation:

Half reaction: $MnCO_{3}\rightarrow MnO_{2}$

(1) $CO_{3}$ balance: $MnCO_{3}\rightarrow MnO_{2}+HCO_{3}^{-}$

(2)H and O balance in acidic medium: $MnCO_{3}+2H_{2}O\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

(3) charge balance: $MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

Hence balanced half-reaction: $MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}$

7 0

3 years ago

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What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm

lesya [120]

Answer: The mass of sucrose required is 69.08 g

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL (Density of water = 1 g/mL)

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

$8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g$

Hence, the mass of sucrose required is 69.08 g

5 0

3 years ago

What is the mass of 0.5 moles of NaCl

Debora [2.8K]

Answer:

29.25g

Explanation:

m=0.5×(23+35.5)

6 0

2 years ago

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Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or $OH^-$ added to the original buffer = 0.05(15.0) = 0.75

This $OH^-$ reacts with HA to form $A^-$ as:

$HA+OH^-\rightarrow H_2O+A^-$

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of $A^-$ after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

$pH=4.20+log(\frac{3.25}{4.25})$

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0

3 years ago

What are downbursts?

katovenus [111]

Downbursts are damaging winds localized in an area which are caused by air that flows downward rapidly out from a thunderstorm. They can cause severe damage to infrastructures and can cause casualties when not prepared. This becomes worse as these types penetrate in areas near the ground.

3 0

3 years ago