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3 years ago

The accepted value is 1.43. Which correctly describes this student’s experimental data?

Chemistry

2 answers:

Yuliya22 [10]3 years ago

7 0

Answer: neither accurate nor precise

Explanation:

Accuracy is defined as how close the measured value is to a standard value. For example if the given volume of water is 20 ml and the two measured values are 19 ml and 18 ml then the former measured value is more accurate than the later.

Precision is defined as how measured values are close to each other. For example, if the length of the wire of 10 m. If the first person measures the length of the same wire thrice and got values 9.7, 9.8 and 9.75 m whereas the second person got values 9.5 , 9.6 and 9.8. In such a case the first person’s measured value is more precise.

Thus as the accepted value is 1.43 and measured values are 1.29 , 1.93 and 0.88 , the experimental data is neither accurate nor precise.

Natalija [7]3 years ago

4 0

Answer:

Neither accurate nor precise

Explanation:

The values were not near or even the same as the accepted value thus making it neither accurate nor precise.

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which one is correct:. . A)Na+ is smaller than K+. B)K+ is smaller than Na+. C)As3- is smaller than Sb3-. D)Sb3- is smaller than

OlgaM077 [116]

Option A is right answer that is Na is smaller than K.

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4 years ago

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A solution is prepared by diluting 50.00 ml of 2.575 m solution of hno3 to 250.0 ml. what is the molarity of the resulting solut

larisa [96]

Thank you for posting your question here at brainly. Below are the choices that can be found elsewhere:

12.88 M
0.1278 M 
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Below is the answer:

5 times diluted (250/50),so 2.575/5=0.515 M

I hope it helps.

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3 years ago

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A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden

Luda [366]

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$ reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

3 0

4 years ago

For the following reaction if you have 13.2g of CO and 42.7g of Fe2O3, which is the limiting reagent with regards to the Fe prod

MrRissso [65]

Answer: CO is a limiting reagent with regards to the Fe production.

Explanation:

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)$

Moles of CO = $\frac{\text{mass of CO}}{\text{molar mass of CO}}=\frac{13.2 g}{28 g/mol}=0.4714 mol$

moles of $Fe_2O_3=\frac{\text{mass of} Fe_2O_3}{\text{molar mass of}Fe_2O_3}=\frac{42.7 g g}{159.7 g/mol}=0.2673 mol$

According to reaction , 3 mole of CO reacts with 1 mole of $Fe_2O_3$ then , 0.4714 moles of CO will react with : $\frac{1}{3}\times 0.4714$ moles of $Fe_2O_3$ that is 0.1571 moles.

0.4714 moles of CO will react with 0.1571 moles of $Fe_2O_3$ which means that CO is present in limited amount acting as limiting reagent.

Mole remaining of $Fe_2O_3$ = 0.2673 mol - 0.1571 mol = 0.1102 mol

Hence, CO is a limiting reagent and $Fe_2O_3$ is an excessive reagent.

8 0

4 years ago

If 90.0 grams of ethane reacted with excess chlorine,how many grams of dicarbon hexachloride would form

tigry1 [53]

Answer:

709 g

Step-by-step explanation:

a) Balanced equation

Normally, we would need a balanced chemical equation.

However, we can get by with a partial equation, as log as carbon atoms are balanced.

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 30.07 236.74

C₂H₆ + … ⟶ C₂Cl₆ + …

m/g: 90.0

(i) Calculate the moles of C₂H₆

n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)

= 2.993 mol C₂H₆

(ii) Calculate the moles of C₂Cl₆

The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)

n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)

= 2.993 mol C₂Cl₆

(iii) Calculate the mass of C₂Cl₆

m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)

m = 709 g C₂Cl₆

The reaction produces 709 g C₂Cl₆.

6 0

3 years ago