Write any three acids which are used in our daily life. Also, write an application of each

At 1.00 atm and 0 ∘ C, a 5.04 L mixture of methane ( CH 4 ) and propane ( C 3 H 8 ) was burned, producing 15.0 g CO 2 . What was

diamong [38]

Answer : The mole fraction of methane and propane is, 0.742 and 0.26

Explanation :

First we have to calculate the moles of mixture by using ideal gas equation.

PV = nRT

where,

P = pressure of the mixture = 1.00 atm

V = Volume of the mixture = 5.04 L

T = Temperature of the mixture = $0^oC=[0+273]K=273K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of mixture = ?

Putting values in above equation, we get:

$1.00atm\times 5.04L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\n_{mix}=\frac{1.00\times 5.04}{0.0821\times 273}=0.225mol$

Let the number of moles of methane be 'x' moles and that of propane be 'y' moles

So, $x+y=0.225$ .....(1)

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, 'x' moles of methane will produce = $\frac{1}{1}\times x=x$ moles of carbon dioxide

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane produces 3 mole of carbon dioxide

So, 'y' moles of propane will produce = $\frac{1}{1}\times y=y$ moles of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

Total moles of carbon dioxide = (x + 3y)

Mass of carbon dioxide = (Total moles) × (Molar mass of carbon dioxide)

Molar mass of carbon dioxide = 44 g/mol

Mass of carbon dioxide = $(x+3y)\times 44$

As we are given:

Mass of carbon dioxide = 15.0 g

So, $44(x+3y)=15.0$ .....(2)

Putting value of 'x' from equation 1, in equation 2, we get:

$44(0.225-y+3y)=15.0\\\\0.225+2y=0.341\\\\y=0.058$

Evaluating value of 'x' from equation 1, we get:

$x+0.058=0.225\\x=0.167$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For Methane:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of methane = 0.167 moles

Total moles = 0.225

Putting values in above equation, we get:

$\chi_{(Methane)}=\frac{0.167}{0.225}=0.742$

For Propane:

Moles of propane = 0.058 moles

Total moles = 0.225

Putting values in above equation, we get:

$\chi_{(Propane)}=\frac{0.058}{0.225}=0.26$

Hence, the mole fraction of methane and propane is, 0.742 and 0.26

6 0

4 years ago

determine the concentration of 35.0mL of an acid solution that requires 25.1mL of a 0.50M NaOH to neutralize it

Degger [83]

Hello!

The general chemical equation for the neutralization of an acid HA with NaOH is the following:

HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

For determining the concentration of the acid solution, we can use the equation shown below:

$moles HA=moles NaOH \\ \\ cHA*VHA=cNaOH*VNaOH \\ \\ cHA= \frac{cNaOH*vNaOH}{VHA}= \frac{0,5M*25,1 mL}{35 mL}=0,36 M$

So, the concentration of the Acid is 0,36 M

Have a nice day!

7 0

4 years ago

For elements 1–36, there are two exceptions to the filling order as predicted from the periodic table. draw the atomic orbital d

boyakko [2]

Here we have to draw the atomic orbital diagram of the elements present in the range of ¹H to ³⁶Kr having exception and its number of unpaired electrons present in the molecule.

The two exceptional elements are chromium (Cr) and Copper (Cu) the atomic orbital diagrams are-

²⁴Cr = 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.

²⁹Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹.

The number of unpaired electrons in Cr and Cu are 6 and 1 respectively.

The atomic orbital diagram of the elements in periodic table is governed by the three rules or principles. Those are (i) The Aufbau principle, (ii) The Pauli exclusion principle and (iii) Hund''s rule of maximum multiplicity.

Among it the Aufbau principle explain the energy level of the atomic orbitals 1s < 2s 2p < 3s < 3P < 3d < 4s etc.

The Pauli's principle tells number of maximum electrons can be present in a particular orbital. As per the principle the s, p and d orbital can posses maximum 2, 6 and 10 electrons.

the Hund's rules explains the extra stability of the half filled and full filled orbitals.

In ²⁴Cr (after ²³V: 1s²2s²2p⁶3s²3p⁶3d³4s²) the expected electronic configuration is 1s²2s²2p⁶3s²3p⁶3d⁴4s² having four unpaired electron but as the half filled d-orbital (5 electrons) is more stable thus it posses 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ electronic configuration having 6 unpaired electrons.

In ²⁹Cu (after ²⁸Ni: 1s²2s²2p⁶3s²3p⁶3d⁸4s²) the expected electronic configuration is 1s²2s²2p⁶3s²3p⁶3d⁹4s² having 1 unpaired electron in d orbital, but due to high stability of the fulfilled d orbital (10 electrons) the atomic orbital diagram become 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ having same number of unpaired electron is s orbital.

The two exceptional atomic orbital diagram of the element in between ¹H to ³⁶Kr is drawn and number of unpaired electrons are calculated.

6 0

3 years ago

10pt if given the correct answer I will mark you as brainliest

diamong [38]

Answer:

C

Explanation:

4(Al)+3(O2)==2(Al2O3)

Also, i'm in sixth grade just took a leson on a text book so may be wrong but i'm prety sure it's the correct nswer because x(Al)+y(O2) can not make hydregen for the h2o and for b, al+so2 = al2O3 makes sense but where did the sulfer go?

3 0

3 years ago

Read 2 more answers

Which of the following minerals is a ferromagnesian silicate?

lutik1710 [3]

Answer: Hornblende

Explanation: Even more permissive than pyroxene, amphibole has compositions that can be extremely intricate. For instance, the elements sodium, potassium, calcium, magnesium, iron, aluminum, silicon, oxygen, etc. can be found in hornblende.

3 0

1 year ago

Write any three acids which are used in our daily life. Also, write an application of each​

Write any three acids which are used in our daily life. Also, write an application of each