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3 years ago

15

Help me Pleaseeeeeeeeeee

1 answer:

Fynjy0 [20]3 years ago

6 0

It’s a synthesis reaction, which is when 2 or more substances react to form a single product.

I hope this helps

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

3 years ago

For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the

Aleks04 [339]

Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

<em>Where A is the absotbance of the solution: 0.216</em>

<em>E is the extinction coefficient = 18000M⁻¹cm⁻¹</em>

<em>b is patelength = 1cm</em>

<em>C is concentration of the solution</em>

<em />

Replacing:

0.216 = 18000M⁻¹cm⁻¹*1cm*C

<h3>1.2x10⁻⁵M = Concentration of the product released</h3>

4 0

3 years ago

3. Ammonia, NH3, is a typical ingredient in household cleaners. It is produced through a combination reaction involving N2(g) an

hram777 [196]

N2 + 3H2 ----> 2NH3

<span>you can see 3 moles H2 reacts to form 2 moles NH3 </span>
<span>Therefore moles NH3 = 2 / 3 x moles H2 </span>
<span>= 2/3 x 12.0 mol </span>
<span>= 8.00 mol NH3 hope this help</span>

6 0

3 years ago

Read 2 more answers

A hydrocarbon sample was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.00 kg of water rose from 20.

fomenos

Answer:

The heat released by the combustion is 20,47 kJ

Explanation:

Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:

Q = C×m×ΔT + Cc×ΔT

Where:

Q is the heat released

C is specific heat of water (4,186kJ/kg°C)

m is mass of water (1,00kg)

ΔT is temperature change (23,65°C - 20,45°C)

And Cc is heat capacity of the calorimeter (2,21kJ/°C)

Replacing these values the heat released by the combustion is:

<em>Q = 20,47 kJ</em>

6 0

3 years ago