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3 years ago

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given the following reaction: CuO (s) + H2 (g) to Cu(s) + H2O (g). If 250 L of hydrogen gas are used to reduce copper (II) oxide

at STP, what mass of copper is to be expected?

1 answer:

QveST [7]3 years ago

4 0

24g is needed no explanation

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What is a centrifuge used for?

tatiyna

<span>It is used for the separation of fluids, gas or liquid, based on density.

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3 years ago

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In order for a process to be spontaneous, Multiple Choice the entropy of the surroundings must increase. the entropy of the surr

Crazy boy [7]

Answer:

The correct answer is entropy change of the surrounding plus the entropy change of the system must be positive.

Explanation:

The term entropy is a state function.Entropy can be defined as the disorder or randomness of the molecules in a system.

A spontaneous reaction is a type of reaction which deals with the release of free energy.The change of free energy in case of spontaneous reaction is always negative.

According to the second law of thermodynamics a spontaneous reaction will occur in a system if the total entropy of both system and surrounding increases during the reaction.

8 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

An organic compound containing C,H and O only. Has 29.6% O by mass. Its relative .olecular mass is 270. How many Oxygen atoms ar

Ostrovityanka [42]

Answer:

12.1% C, 16.1% O, 71.8% Cl

Explanation:

i hope this is correct

6 0

2 years ago

The gas phase reaction, A2 + B2 ---->2AB, proceeds by bimolecular collisions between A2 and B2 c,o(\'e'9 molecules. The rate

Varvara68 [4.7K]

Answer:

The reaction rate becomes quadruple.

Explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.

Thus,

Given that:- The rate law is:-

$r=k[A_2][B_2]$

Now, $[A'_2]=2[A_2]$ and $[B'_2]=2[B_2]$

So, $r'=k[A'_2][B'_2]=k\times 2[A_2]\times 2[B_2]=4\times k[A_2][B_2]=4r$

<u>The reaction rate becomes quadruple.</u>

5 0

3 years ago