Which one of the following is a vestigial structure? 1. Heart 2. Appendix 3. Right leg 4. liver

Change the coefficients until the reaction is balanced. (also if anyone could help me find the rest of the gizmos I would be ver

Artemon [7]

Answer:

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 2NaC₂H₃O

Explanation:

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 1NaC₂H₃O₂

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 2NaC₂H₃O₂

(on the reactant side there are two molecules of C₂H₃O₂ and 2 atoms of Na, so in order to balance that we put the coefficient 2 on NAC₂H₃O₂)

So the final product is

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 2NaC₂H₃O₂

6 0

3 years ago

What is bungee gum made of? A) The properties of both rubber and gum. B) Bungee cord and gum.

slamgirl [31]

The anime character?

5 0

3 years ago

Read 2 more answers

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

VladimirAG [237]

<u>Answer:</u>

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

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

<u>For 3:</u> The equilibrium pressure of oxygen gas is 0.0577 atm

<u>Explanation:</u>

<u>For 1:</u>

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

<u>For 2:</u>

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 $1.386\times 10^{-2}$

<u>For 3:</u>

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

Which car has the most kinetic energy?

luda_lava [24]

Answer:

<h3>.A car of mass 1800kg with speed 15m/s</h3><h3>Opinion C</h3>

7 0

3 years ago

Louie drinks a bottle of a water even though he is not thirsty. As a result, what will the body do to maintain homeostasis?

drek231 [11]

<span>Having too much water in the body could cause so much complications; just as having too little can be dangerous. In each of these scenarios the body tries to maintain equilibrium through a process called homeostasis. Homeostasis depends on many variables, such as body temperature and body fluid being kept within certain pre-set limits. The essence of homeostasis is maintain bodily functions and metabolism at equilibrium state. If Loiue takes in so much water in a very hot atmosphere or during summer even though he's not thirsty, his body would have to get rid of the excess water by sweating. On the other, during winter, or in a cold environment Loiue's body would have to get rid of excess water by urinating frequently.</span>

6 0

3 years ago