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Irina-Kira [14]

4 years ago

9

The accepted value for the molar volume of a gas is 22.4 L. Sarah’s experimental data indicated that a mole of a gas had a volum

e of 19.6L. What was her percent error?

1 answer:

Law Incorporation [45]4 years ago

4 0

Answer:

12.5%.

Explanation:

The following data were obtained from the question:

Accepted value = 22.4 L

Experimental value = 19.6L

Percentage error =.?

Percentage error can be obtained by using the following formula:

Percentage error = |Experimental value – Actual value | / actual value × 100

Percentage error = |19.6 – 22.4|/22.4 × 100

Percentage error = 2.8 /22.4 × 100

Percentage error = 280/22.4

Percentage error = 12.5%

Therefore, the percentage error of Sarah is 12.5%

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A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

Initial mole of Co(NO3)2 $=\frac{mass}{molar mass}$

$=\frac{5.00}{182.94} \\\\=0.02733mol$

Mole of Co(NO3)2 in final solution

$=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol$

Mole of NO3- in final solution = 2 x Mole of Co(NO3)2

$=2\times 0.001093\\\\=0.002186mol$

Mass of NO3- in final solution is mole x Molar mass of NO3

$=0.002186\times62.01\\\\=0.136g$

6 0

4 years ago

Drag the tiles to the correct boxes. Not all tiles will be used. The oxidation number of oxygen is -2, and that of chlorine is -

Evgen [1.6K]

<u>Answer:</u>

<em>That oxidation number of chlorine is -1 while that of oxygen is -2</em>

<u>Explanation:</u>

When it comes to manganese chloride to stabilize the chlorine Ion. Two manganese ions shall be required. Hence the formula for the compound becomes $M_nCl_2$ while the formula for manganese (iv) chloride will be $M_nCl_4$ similarly.

When oxygen and magnesium oxide is considered the Two ions of Manganese will satisfy to ions of oxygen. Which cancels and becomes $M_nO$ while for manganese (iv) oxide the formula becomes $M_nO_2$ .

5 0

3 years ago

Read 2 more answers

The volume of a gas filled balloon is 22L at 313K and 1.5 atm. What would the volume be at 273K and 1 atm?

avanturin [10]

<h3></h3>

From the calculations and the statement of the general gas equation, the volume of the gas becomes 28.8 L

<h3>What are gas laws?</h3>

The gas laws are used to show the relationship between the variables in problems that has to do with gases.

In this case, we have to apply the general gas law as follows;

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = * 22L * 1.5 atm * 273K/313K * 1 atm

V2 = 28.8 L

Learn more about gas laws:brainly.com/question/12667831

#SPJ2

7 0

3 years ago

Read 2 more answers

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH

Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V ) = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI + H₂O ⇄ NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = $\sqrt{Ka.C}$ and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺ in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰

= 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

= 0.359 mol / L * 0.25 L = 0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

= 0.08979 * 53.5 g/mol

= 4.80 g ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

8 0

3 years ago