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3 years ago

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1. What is oxidation? * O a. it is loss of electrons O b. it is gain of electrons O c. it is loss of protons O d. none of the ab

ove

1 answer:

Lemur [1.5K]3 years ago

7 0

Answer:

yes

Explanation:

who wants free points

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N2+3H2 → 2NH3

s2008m [1.1K]

Explanation:

N2 (g) + H2 (g) gives out NH3 (g)

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give NH3 a new stoichiometric coefficient.)

N2 (g) + 3H2 (g) gives out 2NH3 (g)

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one mol of anything is equal in quantity to one mol of anything else because the comparison is made in the units of mols.

So what do we do? Convert to

mols (remember the hint?).

28g N2 × 1 mol N2/ 2 × 14.007) g N2

= 0.9995 mol N2

At this point you don't even need to calculate the number of mols of H2 . Why? Because H2 is about 2 g/mol, which means we have over 10 mols of H2. We have 1 mol N2, and we need three times as many mols of H2 as we have

N2.

After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.

Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:

1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3

= 34.0444 g NH3

Hope it helpz~

4 0

2 years ago

Suppose a new group 5A element is discovered. Photoelectron spectroscopy (PES) reveals that its last two peaks occur at 1370 kJ/

iren2701 [21]

Answer: The average valence electron energy (AVEE) of this element =

1014.2 KJ/ mol or 1.0142mJ/mol.

Explanation:

The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.

The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3

Therefore the average valence electron energy (AVEE) of this element will be calculated as

The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5

2740+2331/ 5 =5071/5

=1014.2 KJ/ mol or 1.0142mJ/mol.

5 0

3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

<u>Answer:</u> The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

3 years ago

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Ivanshal [37]

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

5 0

2 years ago

What is the result of rocks poor ability to transmit heat

Montano1993 [528]

Result rocks poor what heat lave transmit

6 0

3 years ago