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3 years ago

Balance the following equations:

Chemistry

1 answer:

Brilliant_brown [7]3 years ago

6 0

Answer:

1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)

2) V2O5 + CaS -> CaO + V2S5

just additional info: V2O5 is divanadium pentaoxide

LHS (Left hand side)

V: 2

O: 5

Ca: 1

S: 1 x 5 [to balance with the right hand side of the equation]

RHS (Right hand side)

V: 2

O: 1 x 5 [to balance with the left hand side of the equation]

Ca: 1

S: 5

When you balance any elements, you have to balance the whole chemical compound.

Thus,

V2O5 + 5 CaS -> 5 CaO + V2S5

LHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

RHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

3) S8 + O2 -> SO2

LHS:

S: 8

O: 2

RHS:

S: 1 x 8 [to balance with LHS]

O: 2

When you balance any elements, you have to balance the whole chemical compound.

S8 + O2 -> 8 SO2

When we add 8 to the RHS, it gives us 8S, 16 O.

In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.

Therefore, S8 + 8 O2 -> 8 SO2 is the final answer for (3).

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What is the maximum value of the secondary quantum number (l) allowed for atomic electrons with a principal quantum number n = 7

Elis [28]

Explanation:

Principal quantum number=n=7

Azimuthal quantum number=l:-

$\\ \sf\longmapsto n-1$

$\\ \sf\longmapsto 7-1$

$\\ \sf\longmapsto \ell=6$

3 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

Somebody please answer

Zepler [3.9K]

Answer:

177 mmHg

Explanation:

Given that:

The sample of a gas is with Pressure (1) $P_1$ = 500 mmHg

The initial temperature $T_1$ = 773 K

The final temperature $T_2$ = 273 K

Using ideal gas equation:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \\ \\ \dfrac{500}{773} = \dfrac{P_2}{273} \\ \\ (136500) = 773 \times P_2 \\ \\ P_2 = \dfrac{136500}{773} \\ \\ P_2 = 176.58 \ mmHg \\ \\ \simeq 177 mmHg$

7 0

3 years ago

Which of the following is a type of cell organelle?

xz_007 [3.2K]

Answer:

Nucleas is a cell organelle

4 0

3 years ago

Read 2 more answers

A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in

mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is $5.90\times 10^{-4}M$

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock magnesium fluoride solution

$M_2\text{ and }V_2$ are the molarity and volume of chemist's magnesium fluoride solution

We are given:

$M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL$

Putting values in above equation, we get:

$0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M$

Hence, the concentration of chemist's working solution is $5.90\times 10^{-4}M$

8 0

3 years ago