Balance NH4C2H3O2 + PbO2 and tell what type of reaction it is. Please help!! Time is limited!

Periods On the periodic table represent elements

Natalka [10]

Answer:When you look at the periodic table, each row is called a period (Get it? Like PERIODic table.). All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons.

Explanation:

7 0

2 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

in the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2? 2C6H6 + 15O2 → 12CO2 + 6H2O

Mrrafil [7]

Answer: -

12.41 g

Explanation: -

Mass of CO₂ = 42 g

Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol

Number of moles of CO₂ = $\frac{42}{44 g/mol}$

= 0.9545 mol

The balanced chemical equation for this process is

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

From the balanced chemical equation we see

12 mol of CO₂ is produced from 2 mol of C₆H₆

0.9545 mol of CO₂ is produced from $\frac{2 mol C6H6 x 0.9545 mol CO2}{12 mol CO2}$

= 0.159 mol of C₆H₆

Molar mass of C₆H₆ = 12 x 6 + 1 x 6 =78 g /mol

Mass of C₆H₆ =Molar mass x Number of moles

= 78 g / mol x 0.159 mol

= 12.41 g

8 0

3 years ago

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

Katarina [22]

We will use boiling point formula:

ΔT = i Kb m

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass Kg

0.27 = moles / 0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol

5 0

3 years ago

What is the highest point of the wave a transverse wave called

topjm [15]

<span>The answer to the question "what is the highest point of the transverse wave called" is a crest or peak. A transverse wave is a wave in which the medium of the wave vibrates at 90 degrees to the direction in which the wave is moving at. The lowest points are called the troughs. Examples of transverse waves are light and electromagnetic radiation.</span>

7 0

2 years ago

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