Which event always involves a chemical change A boiling B melting C conducting D burning

For the following reaction, calculate how many moles of NO2 forms when 0.356 moles of the reactant completely reacts.

Nesterboy [21]

Answer:

0.712 mol of NO₂ are formed .

Explanation:

For the reaction , given in the question ,

2 N₂O₅ ( g ) → 4 NO₂ ( g ) + O₂ ( g )

From the above balanced reaction ,

2 mol of N₂O₅ reacts to give 4 mol of NO₂

Applying unitary method ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

From the question , 0.356 mol of N₂O₅ are reacted ,

<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>

Since ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

0.356 mol of N₂O₅ reacts to give 4 / 2 * 0.356 mol of NO₂

Calculating ,

0.712 mol of NO₂ are formed .

7 0

2 years ago

Consider the reaction.

N76 [4]

Answer: m = 50 g ZnSO4

Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.

0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

= 0.311 moles ZnSO4

0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4

= 50 ZnSO4

5 0

3 years ago

Read 2 more answers

How is Hydrogen in heavy water different from hydrogen in normal water

soldier1979 [14.2K]

Answer: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.

Explanation:

7 0

2 years ago

The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano

Lubov Fominskaja [6]

Answer:

Ka = 1.52 E-5

Explanation:

CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

3 0

3 years ago

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

2 years ago