Correct possible values of l for n = 4?

How can you distinguish between a physical change and a chemical change?

AleksAgata [21]

Answer:

Physical Change

When a substance undergoes a physical change, its composition remains the same despite its molecules being rearranged.
Physical change is a temporary change.
A Physical change affects only physical properties i.e. shape, size, etc.
A physical change involves very little to no absorption of energy.
Some examples of physical change are freezing of water, melting of wax, boiling of water, etc. .
In a physical change, no new substance is formed.
Physical change is easily reversible i.e original substance can be recovered.

Chemical Change

When a substance undergoes a chemical change, its molecular composition is changed entirely. Thus, chemical changes involve the formation of new substances.
.A chemical change is a permanent change.
Chemical change both physical and chemical properties of the substance including its composition
During a chemical reaction, absorption and evolution of energy take place.
A few examples of chemical change are digestion of food, burning of coal, rusting, etc.
A chemical change is always accompanied by one or more new substance(s).
Chemical changes are irreversible i.e. original substance cannot be recovered.

8 0

3 years ago

Which of the following substance is NOT matter? *

Yuri [45]

All of them are matter tho

5 0

3 years ago

How many significant figures are in 5.40

kotykmax [81]

There are 3 significant figures in this value, all values before and after the decimal point are significant. As there is a decimal point, the zeros trailing are also significant.

4 0

3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

a compound containing only carbon and hydrogen produces 1.80 gCO2 and 0.738 gH2O. Find the empirical formula of the compound. Ex

r-ruslan [8.4K]

Answer:

The empirical formula is CH2

Explanation:

Step 1: Data given

Mass of CO2 produced = 1.80 grams

Mass of H2O produced = 0.738 grams

Step 2: The balanced equation

CxHy + O2 → CO2 + H2O

Step 3: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 1.80 grams / 44.01 g/mol

Moles CO2 = 0.0409 moles

Step 4: Calculate moles C

In 1 mol CO2 we have 1 mol CO2

In 0.0409 moles CO2 we have 0.0409 mol C

Step 5: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 0.738 grams / 18.02 g/mol

Moles H2O = 0.04095 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.04095 moles H2O we have 2*0.04095 = 0.0819 moles H

Step 7: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.0409 / 0.0409 = 1

H: 0.0819 moles / 0.0409 = 2

The empirical formula is CH2

4 0

3 years ago