What is the mass of 2.25 mol of the element Iron (Fe)?

Analyze: The first shell can hold a maximum of two electrons. How does this explain the valence of hydrogen

chubhunter [2.5K]

Answer:

See explanation

Explanation:

Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.

The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).

The formation of the hydride ion completes the 1s orbital.

4 0

2 years ago

A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of

Mars2501 [29]

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $\frac{12}{44}\times 0.3986=0.1087g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $\frac{2}{18}\times 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%$

For Hydrogen:

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0

2 years ago

The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C

Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $460^oC$ = $5.8\times 10^{-6}s^{-1}$

$K_2$ = rate constant at $T_2$ = $4\times K_1$

$Ea$ = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $460^oC=273+460=733K$

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get:

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

$T_2=791.1K$

Therefore, the final temperature would be, 791.1 K

4 0

2 years ago

What type of chemical reaction is this? Cl2(g) + 2KBr(aq) - 2KCl(aq) + Br2(l)

Sindrei [870]

C. Single-replacement

Chlorine replaces Bromine in KBr.

7 0

2 years ago

Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

Regards!

5 0

2 years ago