All categories

3 years ago

3.06: Laboratory: Titration

Chemistry

1 answer:

DENIUS [597]3 years ago

6 0

Answer:

it would be 0.341 because if you add 0.229 and 0.112 it will be 0.341

Explanation:

81.8 g/mol

We’re being asked to calculate the molar mass of an unknown acid based on our interpretation of the titration curve.

Recall that at the equivalence point of a titration:

You might be interested in

Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:

Drupady [299]

Answer:

$r = k [A]^{2}[B]^{2}$

Explanation:

A + B + C ⟶ D

$\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}$

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

$\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}$

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

$\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}$

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

$\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}$

5 0

3 years ago

What is 3623÷72×2(59)1(4)<8335×(3)(2x)

sineoko [7]

Answer:193

Explanation:

7 0

3 years ago

100 ml of soft drink A contains 75g of sugar and weighs 110g. calculate density

a_sh-v [17]

I dont know the answer to this

4 0

3 years ago

You have a 1.8 M solution of HCI. You need to make a 0.70 M solution with a volume of 550 mL. How many mL of the stock solution

Arte-miy333 [17]

The volume of the stock solution needed is 213.88 mL to get new concentration.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Determination of the volume of stock solution.

Volume of diluted solution (V₂) =550 mL

Molarity of diluted solution (M₂) =0.70 M

Molarity of stock solution (M₁) = 1.8 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

1.8 M × V₁ = 0.70 M × 550 mL

V₁ = 213.88 mL

Thus, the volume of the stock solution needed is 213.88 mL.

Learn more about the molarity here:

brainly.com/question/2817451

#SPJ1

5 0

2 years ago

PLEASE HELP ME

ki77a [65]

Answer:

1.15 atm

Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide

We were given the following:

Total pressure = 2.45 atm

Pressure of oxygen = 0.65 atm

Pressure of carbon monoxide = x

Pressure of carbon dioxide = 0.65 atm

Therefore:

2.45 = x + 0.65 + 0.65

2.45 = x + 1.3

x = 2.45 - 1.3

x = 1.15 atm

6 0

4 years ago