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asambeis [7]
2 years ago
11

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

to the actual value, and calculates a 25% error. Assuming she has performed all of the calculations correctly, name one aspect of her methods that may have led to such a high error.
Chemistry
1 answer:
Viktor [21]2 years ago
4 0

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

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Convert 43 F to Kevin. (Remember units on your answer and SHOW YOUR WORK)
gavmur [86]

Answer:

279.26K

Explanation:

T(k) = (43 + 459.67)×5/9

502.67 × 5/9

279.26 K

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Explanation:

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6 0
3 years ago
How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?
Nina [5.8K]

11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2  gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=298 K

Putting value in the given equation:

\frac{PV}{RT}=n

n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}

Moles = 0.3588 moles

Now,

Moles = \frac{mass}{molar \;mass}

0.3588 moles = \frac{mass}{32}

Mass= 11.48 gram

Hence, 11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2 gas measured at 1.3 atm pressure and 298 K

Learn more about the ideal gas here:

brainly.com/question/27691721

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3 0
2 years ago
How should the enthalpy of an intermediate step be manipulated when used to produce an overall chemical equation?
Lapatulllka [165]
<span>The enthalpy of an intermediate step when used to produce an overall chemical equation should be manipulated in this way:
</span><span>Multiply the enthalpy by –1 if the chemical equation is reversed.
If the forward reaction requires energy, the reverse will produce energy.</span>
8 0
3 years ago
Read 2 more answers
A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0
3 years ago
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