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3 years ago

Mass of 0.432 moles of C8H9O4?

2 answers:

7 0

Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169

1 mol = 169 grams.
0.432 mol = x

1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams

romanna [79]3 years ago

4 0

Explanation:

It is known that number of moles present in a substance is equal to the mass divided by molar mass.

Mathematically, No. of moles = $\frac{mass}{\text{molar mass}}$

As it is given that number of moles are 0.432 moles and molar mass of $C_{8}H_{9}O_{4}$ is 169.15 g/mol.

Hence, calculate the mass of $C_{8}H_{9}O_{4}$ is as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.432 mol = $\frac{mass}{169.15 g/mol}$

mass = 73.072 g

Thus, we can conclude that the mass of 0.432 moles of $C_{8}H_{9}O_{4}$ is 73.072 g.

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Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

3 years ago

Include two kinds

MatroZZZ [7]

Answer:

Explanation:

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element.

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are joint together by electrostatic interaction and ionic compound sodium chloride is formed.

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive and both bonded atoms connected together through covalent bond.

8 0

3 years ago

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has $SP^{2}$ O has $SP^{2}$ .

Explanation : The orbitals of oxygen and carbon which are involved in $SP^{2}$ hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and O also has π orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b: O $SP^{3}$ H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen $SP^{3}$ hydrogen atoms in it. It has $SP^{3}$ hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c: C is $SP^{3}$ O is also $SP^{3}$

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of $SP^{3}$ hybridized bonds.

Answer 5) Bond d: C atom has $SP^{3}$ C atom has $SP^{3}$

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be $SP^{3}$ . Therefore, the orbitals that undergo $SP^{3}$ hybridization are $SP^{3}$ .

Answer 6) Bond e : C1 containing O $SP^{2}$

C2 is $SP^{3}$

Explanation : The carbon atom which contains oxygen along with a double bond has $SP^{2}$ hybridized orbitals involved in the bonding process; whereas the carbon at C2 has $SP^{3}$ hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0

3 years ago

I get 2 marks for this so plz answer. How many particles are in a 53.2 L volume of gas at STP?

exis [7]

The number of particles= 1.43 x 10²⁴ particles

<h3>Further explanation</h3>

Given

53.2 L volume of gas at STP

Required

The number of particles

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

So mol for 53.2 L :

= 53.2/22.4

= 2.375 moles

1 mol = 6.02 x 10²³ particles (molecules, atoms, ions)

The number of paricles for 2.375 moles gas :

= 2.375 x 6.02 x 10²³

= 1.43 x 10²⁴ particles

6 0

3 years ago

can anyone help me in explaining the metal excess defect in Non stoichiometric defects? i fail to understand it

mars1129 [50]

<span>Non-stoichiometric defects are </span>compounds which contain the combining elements in a ratio different from that required by their stoichiometric formula. The solids with metal excess <span>defect </span>contain metal in excess to the stoichiometric ratio. Such defect is caused due to either of the following reasons:
1. <span>Metal excess Defect due to Anionic Vacancies:
In this, </span>negative ions may be missing from their lattice sites leaving holes in which the electrons remain entrapped to maintain the electrical neutrality.
2. Metal excess defect due to the presence of extra cations at interstitial sites:
In this case, there are extra positive ions occupying interstitial sites and the electrons in another interstitial sites to maintain electrical neutrality. The defect may be visualized as the loss of non-metal atoms which leave their electrons behind. The excess metal ions occupy interstitial positions.

8 0

3 years ago