Solve the following: Which one is more metallic

Which unit would be used in energy calculations in thermodynamics?

prisoha [69]

Answer:

kJ/mol

Explanation:

4 0

3 years ago

The progress of the reaction:

Blababa [14]

Answer:

The Equilibrium constant K is far greater than 1; K>>1

Explanation:

The equilibrium constant, K, for any given reaction at equilibrium, is defined as the ratio of the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.

It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.

6 0

3 years ago

What is temperature?

BabaBlast [244]

the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch. So I would have to go with A.

8 0

3 years ago

Read 2 more answers

State Boyle's, Charles's, and Gay-Lussac's laws using sentences, then equations

dimaraw [331]

Explanation:

1. Boyle's Law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1\times {V_1}=P_2\times V_2$

2. Charles' Law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

3. Gay Lussac's Law states that tempertaure is directly proportional to the pressure of the gas at constant volume and number of moles of gas

$P\propto T$ (At constant volume and number of moles)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

7 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago