the results have been archived so far in relation to the problem with the given fact relevant and measurable

Calculate the mass of Iron in 40g of Fe203

Nat2105 [25]

Answer:

‏‏‎ ‎

Explanation:‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎

7 0

2 years ago

Scientists have categorized trees based on whether they keep or lose their leaves each year. Another logical way to categorize t

Zolol [24]

Answer:

D

Explanation:

I think but it is an better attempt than the other guy answer.

5 0

2 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago

in an experiment magnesium ribbon was heated in air the product formed was found heavier than original ribbon.potassium manganat

AlladinOne [14]

Answer:

1: due to difference in reactivity and melting point

2 during the reaction both combine with oxygen in this case magnesium remain solid but some potassium vaporise

5 0

3 years ago

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

(b) Moles of $H_2SO_4$ can be calculated as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

Moles of $H_2SO_4$ = 0.00145 moles

From the reaction,

1 mole of $H_2SO_4$ react with 2 moles of NaOH

0.00145 mole of $H_2SO_4$ react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M

7 0

3 years ago

the results have been archived so far in relation to the problem with the given fact relevant and measurable​

the results have been archived so far in relation to the problem with the given fact relevant and measurable