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Determine the Mass Number of the Following Elements:

Phantasy [73]

Answer:

Al 28

B 11

C 14

Sc 46

Na 24

Explanation:

5 0

3 years ago

Fill in the coefficients that will balance the following reaction:

Alex17521 [72]

Answer:

To balance a reaction, the amount of reactants must be equal to the amount of products, as stated by the Law of Conservation of Matter. It may help you to keep track of the number of each element in a list as you try to balance. It's not able to be balanced.

3 0

3 years ago

Read 2 more answers

Calculate the acceleration of a car if its speed increases from 4.0 m/s to 16 m/s in 3.7 s.

ioda

Answer:

3.24m/s^2

Explanation:

The following data were obtained from the question:

U = 4m/s

V = 16m/s

t = 3.7secs

Acceleration is change in velocity with time i.e

a = (V — U) / t

a = (16 — 4) / 3.7

a = 12/3.7

a = 3.24m/s^2

8 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

Balance the following equation on scrap paper:<br> AIF3+Li2O → Al2O3 + LiF

Oxana [17]

Answer:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Explanation:

AlF₃ + Li₂O —> Al₂O₃ + LiF

The above equation can be balanced as follow:

AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by writing 2 before AlF₃ as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 6 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 6 before LiF as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + 6LiF

There are 2 atoms of Li on the left side and 6 atoms on the right side. It can be balance by writing 3 before Li₂O as shown below:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Thus, the equation is balanced..!

8 0

3 years ago