What would be the formula for do diphosphorus pentoxide

In the Haber Process, hydrogen (H2) and nitrogen (N2) react to produce

lesya692 [45]

Answer:

Mass stays the same because no matter is created or destroyed.

Explanation:

Regardless of what chemical reaction we have, in each case the law of mass conservation applies. The law of mass conservation states that the total mass of a reaction mixture is kept constant, as mass cannot be created or destroyed.

In this specific reaction, the total mass of the reactants should be equal to the total mass of the products when the reaction is complete.

In other words, if we add the mass of hydrogen to the mass of nitrogen, when the reaction is compete, assuming no reagent in excess, this should be equal to the mass of ammonia formed.

7 0

2 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

A glass jar is sealed, trapping air inside. The jar warms up after it is left ourside in the sun. What will happpen to the press

amid [387]

Answer:

the particles start to spread out and cause it to get warmer

Explanation:

7 0

2 years ago

What effects do melatonin’s have

nataly862011 [7]

Answer: Melatonin has been used safely for up to 2 years in some people. However, it can cause some side effects including headache, short-term feelings of depression, daytime sleepiness, dizziness, stomach cramps, and irritability. Do not drive or use machinery for four to five hours after taking melatonin.

7 0

2 years ago

1. Ethanol and 2,6-dimethylphenol both have hydroxyl groups that can be deprotonated by a base. Why does ethanol act as a solven

Volgvan

Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.

<h3>Ethanol is either a nucleophile or a base.</h3>

The ethanol is a base Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.

<h3>How do solvents impact anionic nucleophile's reactivity?</h3>

In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.

Learn more about nucleophiles here:-

brainly.com/question/27127109

#SPJ4

3 0

1 year ago