Classify the following as an element, compound, homogeneous mixture, or heterogeneous mixture:

Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

$2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}$

Converting moles of boric acid to mass:

$0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g$

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.

5 0

2 years ago

Daniel is developing a model of the light-independent reactions of photosynthesis, also known as the Calvin cycle. How should th

marshall27 [118]

Answer:

*The model should show the carbon compounds enter as carbon dioxide

*The model should show the carbon compounds exit as 3-carbon molecules

Explanation:

In plants, carbon dioxide (CO2) enters the chloroplast through the stomata and diffuses into the stroma of the chloroplast—the site of the Calvin cycle reactions where sugar is synthesized. The reactions are named after the scientist who discovered them, and reference the fact that the reactions function as a cycle.

8 0

2 years ago

We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?

Anna11 [10]

M=D*V
D=620 g/cm³
V=75 cm³

m= 620 g/cm³ * 75 cm³=46500 g
m=46500g

8 0

2 years ago

Several amino acids are intermediates of the urea cycle, having side ammonia groups that join with free carbon dioxide (CO2) and

mr_godi [17]

Answer:

A. Arginine

Explanation:

The urea cycle is the cycle of the biochemical reactions which produces urea from ammonia.

Steps of the urea cycle:

Carbamoyl phosphate, in presence of ornithine transcarbamoylase is converted to citrulline by the denotation of carbamoyl phosphate groupto ornithine and a phosphate group is released.
Amino group of the aspartate and carbonyl group of the citrulline are condensed to form argininosuccinate in the presence of enzyme, argininosuccinate synthetase. This condensation reaction is ATP dependent.
Argininosuccinate then undergoes cleavage by the argininosuccinase to form intermediate, arginine and fumarate.
Arginine is then cleaved by the arginase to form urea as well as ornithine. Ornithine is transported back to mitochondria to begin urea cycle again.

7 0

2 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

Answer: The above reaction is non-spontaneous.

Explanation:

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0

2 years ago