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3 years ago

How do I find the moles of OH- which reacted (mol) in the titration. Table Attached

Chemistry

1 answer:

BaLLatris [955]3 years ago

4 0

Answer:

It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.

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A car tire has a pressure of 2.38 atm at 15.2 c. If the pressure inside reached 4.08 atm, the tire will explode. How hot would t

stich3 [128]

The correct answer is 221.06 °C hot.

If P₁ is the pressure at T₁ and P₂ is the pressure at T₂ then,

P₁/T₁ = P₂/T₂

It is given that P₁ = 2.38 atm

T₁ = 15.2 degree C = 273 + 15.2 = 288.2 K

P₂ = 4.08 atm

T₂ = x

Thus, 2.38 / 288.2 = 4.08 / x

x = (4.08 × 288.2) / 2.38

x = 494.06 K

x = 494.06 - 273 °C = 221.06 °C

Therefore, the tire would get 221.06 °C hot.

8 0

3 years ago

(02.03 LC)

OLga [1]

The age of earth I think

3 0

3 years ago

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What is the specific name for the electrons involved in bonding?

Mariana [72]

Answer:

valence electrons are the ones

4 0

3 years ago

A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97. ??. Unfortunately, a car

serg [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Answer: 760 uM

Explanation:

the addition of solvent to a solution in a such away that the volume of the solution increases and the concentration decreases is a process know as dilution

The concentration and the volume of the dilute and concentrated solution are given below as follows

$C_{1}V_{1}$ $= C_{f}V_{f}$

Concentrated solution Dilute solution

Given that M1 = 97.0uM

V1 = 47.0 mL

V2 = 6.0 mL

M2 = ?

Therefore M1V1 = M2V2

M2 = M1V1/V2

M2 = (97*47)/6

M2 = 760 uM

Answer: 760 uM

5 0

4 years ago

Is first order in bro3⎻ , second order in br⎻, and zero order in h+. By what factor will the reaction rate change if the concent

nata0808 [166]

For a general reaction,

$A+B\rightarrow C$

General expression for rate law will be:

$r=k[A]^{a}[B]^{b}$

Here, r is rate of the reaction, k is rate constant, a is order with respect to reactant A and b is order with respect to reactant B.

The reaction is first order with respect to $BrO_{3}^{-}$ , second order with respect to $Br^{-}$ and zero order with respect to $H^{+}$ .

According to above information, expression for rate law will be:

$r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}$

Or,

$r=k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (1)

When concentration of $BrO_{3}^{-}$ get doubled, rate of the reaction becomes,

$r^{'}=2k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (2)

Dividing (2) by (1)

$\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2$

Or,

$r^{'}=2r$

Thus, rate of the reaction also get doubled.

When the concentration of $Br^{-}$ is halved, the rate of reaction becomes

$r^{"}=k[BrO_{3}^{-}]([Br^{-}]/2)^{2}$

Or,

$r^{"}=1/4k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (3)

Dividing (3) by (1)

$\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}$

Or,

$r^{"}=\frac{r}{4}$

Thus, rate of reaction becomes 1/4th of the initial rate.

When the concentration of $H^{+}$ is tripled:

Since, the rate expression does not have concentration of $H^{+}$ , it is independent of it. Thus, any change in the concentration will not affect the rate of reaction and rate of reaction remains the same as in equation (1).

7 0

4 years ago