What coefficients are needed to balance the following:

Which products are the result of a neutralization reaction? (1 point)

Nata [24]

Answer:

B) water and salt is the result of neutralization reaction

3 0

3 years ago

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

Pavlova-9 [17]

Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

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

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

8 0

3 years ago

In your reaction you started with a solution that was approximately 8 M acetic acid and 4 M isopentyl alcohol. Use the equilibri

V125BC [204]

The final concentrations of water, isopentyl acetate, isopentyl alcohol, and acetic acid obtained on using equilibrium expression are 3.4M, 3.4M, 0.6M , 4.6M

Using ICE table:-

Acetic acid + Isopentyl alcohol ------> Isopentyl acetate + water

I 8M 4M 0 0

C 8-x 4-x +x +x

E 8-x 4-x +x +x

Equilibrium constant (Keq) = 4.2

On using the equilibrium expression -

Keq = [Product] / [Reactant]

Keq = x^2 / (8-x) (4-x )

4.2 = x^2 / 32 - 8x - 4x + x^2

x = 3.4

Concentration of acetic acid = 8-x = 8-3.4 = 4.6M

Concentration of isopentyl alcohol = 4-x = 4-3.4 = 0.6M

Concentration of isopentyl acetate = x = 3.4M

Concentration of water = x = 3.4M

Learn more about equilibrium expression

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4 0

2 years ago

Given 8.25 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

mash [69]

Answer:

10.87 g of Ethyl Butyrate

Solution:

The Balance Chemical Equation is as follow,

H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O

According to equation,

88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate

So,

8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate

Solving for X,

X = (8.25 g × 116.16 g) ÷ 88.11 g

X = 10.87 g of Ethyl Butyrate

8 0

3 years ago

Which of the following occurs when a large nucleus is split apart?

grin007 [14]

Answer: The correct answer is <u>''fission</u>'' it occurs when a large nucleus is split apart.

3 0

3 years ago