how are aqi and aqhi similar

Explanation:

here's the answer to your question

Explanation:

here's the answer to your question

Decomposition of Sulfurous Acid (H₂SO₃):

H₂SO₃ → H₂O + SO₂

In this reaction, sulfurous acid decomposes into water (H₂O) and sulfur dioxide (SO₂).

The decomposition of carbonic acid (H₂CO₃) and sulfurous acid (H₂SO₃) can be represented by the following balanced chemical equations:

Decomposition of Carbonic Acid (H₂CO₃):

H₂CO₃ → H₂O + CO₂

In this reaction, carbonic acid decomposes into water (H₂O) and carbon dioxide (CO₂).

Decomposition of Sulfurous Acid (H₂SO₃):

H₂SO₃ → H₂O + SO₂

In this reaction, sulfurous acid decomposes into water (H₂O) and sulfur dioxide (SO₂).

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Answer:

three electron shells

14 electrons

14 neutrons

Explanation:

Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.

Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.

The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.

The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

Replacing:

P* 5.005 L= 1.255 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] *273.5 K

Solving:

[tex]P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}[/tex]

P= 5.62 atm

The pressure is 5.62 atm.

Answer:

The answer is provided below

Explanation:

To determine the metal is gold we will use the following steps

Take a full water container

Place the metal in the container

Collect the water that spills out due to the placement of the metal

measure the mass of collected water.

Calculate the value in terms of the density of water, it will be the volume of metal.

Calculate the mass of the metal

Use the following formula to calculate the density of the metal

Density = Mass / Volume

Now compre the resulted density to the density of pure gold.

Answer:

hshssytdtctdyeheb

Explanation:

yye6d66d6d6dududyydydydyehwj2

Answer:

-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly

Answer:

I believe it might be blood exiting the lungs, not positive though.

Explanation:

i think it is blood exiting the lungs

Answer:

121 °C

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 374 °C

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

NOTE: Volume = constant

Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 374 °C

Initial temperature (T₁) = 374 °C + 273

Initial temperature (T₁) = 647 K

Next, we shall determine the final temperature. This can be obtained as follow:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 647 K

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

P₁ / T₁ = P₂ / T₂

5.97 / 647 = 3.64 / T₂

Cross multiply

5.97 × T₂ = 647 × 3.64

5.97 × T₂ = 2355.08

Divide both side by 5.97

T₂ = 2355.08 / 5.97

T₂ = 394 K

Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 394 K

Final temperature (T₂) = 394 – 273

Final temperature (T₂) = 121 °C

Thus, the new temperature is 121 °C

Answer:

The correct solution is "3.32 gm/L".

Explanation:

Given:

Rate constant,

[tex]K = 0.05 \ hr^{-1}[/tex]

Time,

[tex]t = 24 \ hours[/tex]

Concentration of ethanol,

[tex]C_o= 1 \ mg/L[/tex]

Now,

The concentration of ethanol after 24 hours will be:

⇒ [tex]C_o=C\times e^{-K\times t}[/tex]

By putting the values, we get

    [tex]1=C\times e^{-0.05\times 24}[/tex]

    [tex]1=C\times 0.30119[/tex]

    [tex]C= 3.32 \ gm/L[/tex]

Answer:

B

Explanation:

E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.

Answer:

a) You can form 466 g of BaSO₄.

Explanation:

a) Mass of BaSO4

196 g H₂SO4 × 1 mol H₂SO4

98.08 g H₂SO4

1 mol BaSO 1 mol H₂SO4 X X

466 g BaSO4

233.39 g BaSO4

1 mol BaSO4

Answer:

Rock A will have far more chemical weathering than Rock B due to the rise in area effect

Explanation:

Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.

hope it will help you

Answer:

26kJmole,84kJmole

Explanation:

A catalyst lowers the activation energy between reactants and products. Once this energy barrier is lowered, reactants are converted into products faster.

However, the enthalpy change of a reaction is constant both in the catalysed and uncatalysed reaction.

The activation energy of the uncatalysed reaction must be higher than the activation energy of the catalysed reaction. Hence the answer above.

Answer:

the entropy of the universe increases

Explanation:

The second law of thermodynamics can be stated in terms of entropy, in this statement of the law; ''a spontaneous process increases the entropy of the universe''.

Given that; ∆Suniverse = ∆Ssystem + ∆Ssurroundings, the definition of a spontaneous process is one in which ∆Suniverse >0.

Hence, a process is spontaneous when the entropy of the universe increases.

Answer: The molar mass of solute is 115 g/mol.

Explanation:

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]

OR

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]

Boiling point of solution = [tex]81.20^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute = 10 g

[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 200 g

Putting values in equation 1, we get:

[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]

Hence, the molar mass of solute is 115 g/mol.

Answer:

dont buy cheap and off we went

A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium

Explanation:

Ore contains ---- 65.2% [tex]TiO_2[/tex]

Mass% of titanium in TiO2 can be calculated as shown below:

[tex]mass percentage of Ti in TiO2=\frac{mass of Ti}{mass of TiO_2} *100\\=(47.86g/79.866g)* 100\\=59.9[/tex]

Given 10.0 metric tons of titanium is required.

The mass of ore that should be processed can be calculated as shown below:

Mass of Ti = ore x TiO2 % x Ti mass %

10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti

=>M=(10.0 metric tons) / (0.652 x 0.599)

=>M=25.6 metric tons

Hence, the mass of ore required is 25.6 metric tons.

The minimum amount of rutile ore to be processed is 25.6 metric tons.

Explanation:

Given :

The sample of rutile, 65.2% [tex]TiO_2[/tex] (s) by mass.

To find:

The minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium

Solution:

The atomic mass of titanium = 47.867 g/mol

The atomic mass of oxygen = 15.999 g/mol

The molar mass of [tex]TiO_2[/tex] [tex]= 47.867 g/mol+2\times 15.999 g/mol=79.865 g/mol[/tex]

Percentage of the titanium in  [tex]TiO_2[/tex] :

[tex]Ti(\%)=\frac{1\times 47.867 g/mol}{79.865 g/mol}\times 100\\=59.935\%[/tex]

Quantity of titanium required = 10.0 metric ton

Quantity of  [tex]TiO_2[/tex]  from rutile ore = x

[tex]59.935\%=\frac{10.0 \text{metric ton}}{x}\times 100\\x=\frac{10.0 \text{metric ton}}{59.935}\times 100=16.7 \text{metric ton}[/tex]

Mass of [tex]TiO_2[/tex] form rutile ore= 16.7 metric ton

The percentage of [tex]TiO_2[/tex] in rutile = 65.2 %

The quantity of rutile ore to be processed = M

[tex]65.2\%=\frac{16.7 \text{metric ton}}{M}\times 100\\M=\frac{16.7 \text{metric ton}}{65.2}\times 100=25.6 \text{metric ton}[/tex]

The minimum amount of rutile ore to be processed is 25.6 metric tons.

Learn more about the percent of an element in the compound here:

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Answer: [tex]6.276\ kJ[/tex]

Explanation:

It is known that 1 cal is equivalent to 4.184 J

1500 cal will be equivalent to [tex]1500\times 4.184=6276\ J[/tex]

Also, 1 kJ is equivalent to 1000 J

So, 6276 J is equal to [tex]6.276\ kJ[/tex]

Answer:

C) EtOH 1% AgNO3

Answer:

About 3500 grams of tin.

Explanation:

We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.

First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.

By smelting it, we acquire elemental tin and oxygen gas. Hence:

[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]

(Note: oxygen is a diatomic element.)

The equation is balanced as well.

To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.

Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:

[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]

Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:

[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]

We can convert this into moles.

Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:

[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]

With SnO₂ in the denominator to simplify units.

Finally, we can convert from moles Sn to grams Sn using its molar mass:

[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

With the initial value and above ratios, we acquire:

[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

Cancel like units:

[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]

Multiply. Hence:

[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]

Since we should have two significant figures:

[tex]=3500 \text{ g Sn}[/tex]

So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.

Answer:

3546g

Explanation:

start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction

SnO2 -> Sn + O2

Sn molecular wt: 119

O2 molecular wt: 32

SnO2 molecular wt:  119+32 = 151

so Sn / SnO2 wt ratio = 119 / 151

4.5 kilograms of tin (IV) oxide will produce:

= 4.5 * 119 / 151

= 3.546 kg

or 3546 grams of tin metal

no need to involve moles ;)

Answer:

It increases but less than double

Explanation:

As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.

We know, the kinetic energy of an ideal gas is given by :

[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]

where, R = gas constant

            T = absolute temperature

            M = molecular mass of the gas

From the above law, we get

[tex]$V_{avg} \propto \sqrt{T}$[/tex]

Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.

In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]

[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]

Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]

Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.

Thus, the answer is " It increases but less that double".