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3 years ago

Which type of reaction occurs in the following? 2HgO(s) 2Hg(l) + O2(g)

Chemistry

2 answers:

melomori [17]3 years ago

8 0

A.a decomposition reaction

kotykmax [81]3 years ago

5 0

In this reaction, reactant HgO decomposes into more than one products hence reaction is called decomposition.Hg in HgO has oxidation number of +2 whereas oxidation number of Hg in product side is zero( zero because Hg is in its elemental form) so Hg is reduced. Hence A is correct option.

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A student makes two mixtures, both made of the same substances. What must be true of the two mixtures? A Both mixtures have the

Goshia [24]

Answer:

im not completely sure but the mst likely answer would be the would be they have the same density

Explanation:

if two mixture have the same ingriedients their most likely going to have the same density depending of the measurements like when baking a cake when you add your wet ingriedients to your dry one it makes a batter and if you were to make another mixture with the same ingriedients and somewhat similar measurements your going to get a similar density

6 0

2 years ago

Aluminum chloride can be formed from its elements:

saul85 [17]

<u>Answer:</u> The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

2 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

5 0

2 years ago

What is the answers and pls show work if possible!!

taurus [48]

D = m / V

It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...

V = L x W x H

Volume = Length x Width x Height

start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g

V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3

Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm

Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm

Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm

Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.

Atomic radius gold = 174 pm

Diameter = 348 pm

46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold

4 0

3 years ago

the normal human eye responds to visible light of wavelength range from about 390 to 710nm. Determine the frequency range of the

antoniya [11.8K]

Answer:

$\:$

A typical human eye will respond to wavelengths from 390 to 710 nm (in vacuum or air). In terms of frequency this corresponds to a band in the vicinity of 420–790 THz.

3 0

2 years ago

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