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4 years ago

A sentence with unbalanced

1 answer:

8 0

When a weight is reciprocated, the equal and opposite force required for its acceleration at any instant appears as an unbalanced force on the frame of the machine to which the weight belongs.

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago

A 83.2 g sample of metal at 90.31 oC is added to 41.82 g of water that is initially at 19.93 oC. The final temperature of both t

MissTica

Answer: 0.2885J/g°c

Explanation:

Loss of heat of metal = Gain of heat by the water

Therefore

-Qm = +Qw

Where

Q = mΔTCp

Q = heat

M= mass

ΔT = T.f - Ti

Ti= initial temperature

T.f= final temperature

Cp= Specific heat (m is metal, w is water)

-Qm = +Qw

-[m(T.f-Ti)Cpm] = m(T.f-Ti)Cpw

Therefore

-[83.2g(28.42-90.31)Cpm] = 41.82g(28.42-

19.93)4.194J/g°c

Cpm = 1485.54÷ 5149.25

Cpm = 0.2885J/g°c

Therefore , the specific heat of the metal is 0.2885J/g°c

4 0

3 years ago

What is the atomic number of hydrogen

olga nikolaevna [1]

The atomic number of hydrogen is 1

8 0

3 years ago

1.5g of magnesium ribbon is burnt in oxygen to produce 2.5g of magnesium oxide. How much oxygen is required in grams?

Furkat [3]

Answer:

2.5 grams

Explanation:

law of conservation of mass applies. the magnesium completely burns

4 0

3 years ago

PLEASE HELP.!

Agata [3.3K]

Answer:

Explanation:

6 0

3 years ago