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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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When naming compounds part of the second element's name is dropped and what is added in its place?

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Answer:

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Explanation:

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3 years ago

The H⁺ concentration in an aqueous solution at 25 °C is 9.1 × 10⁻⁴. What is [OH⁻]?

Vinvika [58]

Steps:-

First we calculate pH then pOH then [OH-]

$\\ \tt\rightarrowtail pH=-log[H^+]$

$\\ \tt\rightarrowtail pH=-log[9.1\times 10^{-4}]$

$\\ \tt\rightarrowtail pH=-log9.1-log10^{-4})$

$\\ \tt\rightarrowtail pH=0.95+4$

$\\ \tt\rightarrowtail pH=4.95$

Now

$\\ \tt\rightarrowtail pH+pOH=14$

$\\ \tt\rightarrowtail pOH=14-4.95$

$\\ \tt\rightarrowtail pOH=9.05$

$\\ \tt\rightarrowtail -log[OH^-]=9.05$

$\\ \tt\rightarrowtail log[OH^-]=-9.05$

$\\ \tt\rightarrowtail OH^-=10^{-9.05}$

$\\ \tt\rightarrowtail OH^-=8.91\times 10^{-4}$

5 0

2 years ago

At 25 ∘C , the equilibrium partial pressures for the reaction were found to be PA=5.16 bar, PB=5.04 bar, PC=4.11 bar, and PD=4.8

erastova [34]

Answer: 5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

$A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)$

The expression for equilibrium reaction will be,

$K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}$

Now put all the given values in this expression, we get the concentration of methane.

$K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}$

$K_p=10.6$

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

$\Delta G^o=-2.303\times RT\times \log K_c$

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = $25^0C=(25+273)K=298 K$

$K_c$ = equilibrium constant = 10.6

$\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)$

$\Delta G^o=5850.23J/Kmol$

$\Delta G^o=5.85kJ/Kmol$

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

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The answer is compound.

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