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2 years ago

Which is the correct empirical formula for C6H1206?

Chemistry

1 answer:

Bas_tet [7]2 years ago

8 0

Explanation:

the correct empirical formula for C6H12O6 is CH2O...

hope it will help....

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Consider the following reaction:

Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

$SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Initial conc. $2.4\times 10^{-2}$ 0 0

At eqm. $(2.4\times 10^{-2}-x)$ x x

As we are given:

Concentration of $Cl_2$ at equilibrium = $1.3\times 10^{-2}M$

That means,

$x=1.3\times 10^{-2}M$

The expression for equilibrium constant is:

$K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$

Now put all the given values in this expression, we get:

$K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}$

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}$

$K_c=0.0154$

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0

2 years ago

According to the reaction represented by the unbalanced equation above, how many moles of so2(g) are required to react completel

melomori [17]

2SO₂ + O₂ = 2SO₃

n(O₂)=1 mol

n(SO₂)=2n(O₂)

n(SO₂)=2 mol

6 0

2 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2

Marizza181 [45]

Answer:

$\Delta _rH=-1124.14kJ/mol$

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

$2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)$

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

$\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}$

In such a way, by using the NIST database, we find that:

$\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol$

Thus, we plug in the enthalpies of formation to obtain:

$\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol$

Best regards!

8 0

2 years ago

What volume does .0685 mol of gas occupy at STP? 1mole =22.4

GuDViN [60]

V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm

4 0

2 years ago

Read 2 more answers

HELPPPP. SHOW WORK

nekit [7.7K]

I think The answer to two is it would sink

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2 years ago