How many bonding electrons are present in PO43- ?

What amount of carbon dioxide (in moles) is produced from the reaction of 2.24 moles of ethanol with excess oxygen?

algol13

Answer : The amount of carbon dioxide produced is, 197.12 grams.

Explanation : Given,

Moles of ethanol = 2.24 mole

Molar mass of carbon dioxide = 44 g/mole

The balanced chemical reaction will be,

$C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$

First we have to calculate the moles carbon dioxide.

From the balanced chemical reaction, we conclude that

As, 1 mole of ethanol react to give 2 moles of carbon dioxide

So, 2.24 mole of ethanol react to give $2\times 2.24=4.48$ moles of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=4.48mole\times 44g/mole=197.12g$

Therefore, the amount of carbon dioxide produced is 197.12 grams.

8 0

2 years ago

How do clouds and precipitation form

Elza [17]

Umm, it depends on what lesson you are learning- but clouds form bc it’s so cold in the sky that the moisture forms clouds. precipitation is just water that evaporated into the clouds and the clouds hold it til they are full and then the precipitation falls

5 0

3 years ago

Read 2 more answers

The temperature of a sample of water increases from 20c to 46.6c as it absorbs 5650 Joules of heat. What is the mass of the samp

WINSTONCH [101]

Answer: 51 g

Explanation:

5 0

2 years ago

Which has the highest boiling point .33 m NH3 or .10 m Na2SO4

Aleksandr-060686 [28]

Answer:

0.33 mol/kg NH₃

Explanation:

Data:

b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.

1 mol NH₃ ⟶ 1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

5 0

3 years ago

An evaporation-crystallization process is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fres

shtirl [24]

Answer:

From the degree of freedom analysis, the degree of freedom of the system and its components are equal to zero hence the system is well defined

Explanation:

To appraise the evaporation-crystallisation process, we go over the system to check if it is well defined from the available information as follows

45% by weight of inlet water is evaporated hence where inlet consists of 19.6% by weight of K₂SO₄ we have, Molar mass of K₂SO₄ = 174.259g/mol. Thus for every mole of K₂SO₄, we have 174.259×100/19.6 = 889.1g of solution is fed per mole of K₂SO₄,

Also the stream of concentrate leaving the evaporator contains

889.1 – 174.259 = 714.7 grams of water, and if 45% by weight of water is evaporated we have

45% of 889.1 is evaporated leaving a solution of weight = 889.1 × 55/100 = 489grams of solution which contains

100×174.259÷489 or 35.6% by weight of K₂SO₄ concentrate leaving the evaporator and moving on to the evaporator

However, 175. kg of water is evaporated/s hence from the previous calculation, quantity of water per mole of K₂SO₄ evaporated = 0.45×889.1= 400.1g which in comparison with actual quantity gives mass flow into

0.4001/175 = 0.003 or 437.39 to 1 hence the mass flow rate is 889.1g×437.39 = 388884g/s or 388.9Kg/s

a. Degrees of freedom analysis for the overall system

We have the following 4 unknowns in the overall system viz

m1, m3, m4 and m5

where m1 = maximum mass flow rate into the system

m3 = mass rate of evaporated water

m4 = maximum mass of solid K₂SO₄ crystals produced from the crystallizers

m5 = recycle ratio

While we have

1) Information, maximum capacity of evaporation from where we can calculate the maximum rate of feed supply

2) information, including chemical formula, to determine the maximum production rate

3) Information to calculate the water evaporated from fraction of water which is evaporated to that which is supplied

4) information to calculate the recycle ratio

Hence degrees of freedom = 4 – 4 =0

b. Degree of freedom analysis for the recycle-fresh feed mixing point

For the recycle-fresh feed mixing point we have m1 and m5, two unknowns

Where m1 is maximum feed rate and m5 is the mass of filtrate and we are given the compound molecular formula and the maximum flow rate from where we can calculate both m1 and m5

Hence the degrees of freedom = 0

c. Degree of freedom analysis for the evaporator

For the evaporator we have three unknowns m1,m2 and m3 and the available information are

1. The maximum water processing capacity of the evaporators and

2. The percentage quantity of water evaporated

Which is 2 hence we have 2 – 2 = 0 degrees of freedom

and

d. Degree of freedom analysis for the crystallizer the unknowns are m2, m4, m5

For the crystallizer the unknowns are m2, m4, m5The information available are

1. the ratio of crystals per kilogram of solution

2. The concentration of the recycled K₂SO₄ solution

3. Information of the maximum capacity of the evaporator so as to calculate the mass of concentrates leaving the evaporator and moving towards the crystallizer

Hence, we have 3 -3 = 0 degrees of freedom

From the degree of freedom analysis, the degree of freedom of the system and its components are zero hence the system is well defined

3 0

3 years ago