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3 years ago

What are standards or attributes of a design that can be measured.

Chemistry

1 answer:

ankoles [38]3 years ago

4 0

Explanation:

attribute of a person that often cannot be measured directly but can be assessed using numbers of indicators or manifest variables

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A student obtains a sample of a pure solid compound. In addition to Avogadro’s number, which of the following must the student k

katrin2010 [14]

Answer:

Molar mass of the compound, mass of the sample

5 0

3 years ago

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Sodium carbonate can be made by heating sodium bicarbonate: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) Given that ΔH° = 128.9 kJ/m

Ad libitum [116K]

Answer:

[tex]128^{o}C[/tex]minimum temperature will the reaction become spontaneous.

Explanation:

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

From the given,

$\Delta G^{o}=33.1$

$\Delta H^{o}=128.9kJ/mol$

$Temperature\,change=25+273=98K$

$33.1=128.9-298 \Delta S$

$\Delta S=0.322kJ/mol.K$

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

$0=128.9-T(0.322)=401K=218^{o}C$

Therefore, $128^{o}C$ minimum temperature will the reaction become spontaneous.

7 0

3 years ago

Write the name of the first twenty elements with their symbols and their atomic number

7nadin3 [17]

Answer:

1 Hydrogen H

2 Helium He

3 Lithium Li

4 Beryllium Be

5 Boron B

6 Carbon C

7 Nitrogen N

8 Oxygen O

9 Fluorine F

10 Neon Ne

11 Sodium Na

12 Magnesium Mg

13 Aluminium Al

14 Silicon Si

15 Phosphorus P

16 Sulfur S

17 Chlorine Cl

18 Argon Ar

19 Potassium K

20. calcium Ca

6 0

3 years ago

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9. A 35.2-mL, 1.66 M KMnO4 solution is mixed with 16.7 mL of 0.892 M KMnO4 solution. Calculate the concentration of the final so

saw5 [17]

Answer:

I think that the answer is 0.79M

7 0

3 years ago

For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)

ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e $Fe_2O_3$ .

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

$Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)$

Moles of iron(III) oxide : $\frac{53.7 g}{160 g/mol}=0.3356 mol$

Moles of aluminium : $\frac{22.8 g}{27 g/mol}=0.8444 mol$

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

$\frac{2}{1}\times 0.3356 mol=0.6712 mol$ of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e $Fe_2O_3$ and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

$\frac{1}{1}\times 0.3356 mol=0.3356 mol$ of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

7 0

3 years ago