How many moles of magnesium oxide are produced by the reaction of 1.82 g of magnesium nitride with of water

Answer:

Explanation:

ethyl 3-methylbenzoate

Answer:

Octadecanol: C18H380

Dimethyl sulfoxide: CHOS

Explanation:

Covalent compounds are formed between non-metallic elements

here's the answer to your question

Answer:

Hence the evaporation rate in pints per minute is 0.001583.

Explanation:

Now,

1.5 gallons of paint are used.

It has 5% solid, so total 95% of 1.5 gallon can evaporate(only the solvent part will evaporate).

95% of 1.5 gallon=(95/100)*1.5 = 1.425 gallons.

1gallon = 8 pints.

So 1.425 gallons=8*1.425 pints

=11.4 pints.

Evaporation requires 2 hours. That is 3600*2 seconds = 7200 seconds.

So evaporation rate= 11.4 pints/7200 seconds.

=0.001583 pints per second

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 ;)

Explanation:

here is the answer to your question

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.

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]

hope it will help you

Answer:

[tex]V_2=482.5mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law due to the fact that we are dealing with variable volume, temperature and pressure:

[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]

In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]

Regards!

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:

[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to set up the required chemical reaction as follows:

[tex]C_{14}H_8O_2+O_2\rightarrow CO_2+H_2O[/tex]

However, it turns out mandatory to balance it according to the law of conservation of mass, which states that the atoms must the same on the reactants side as on the products side; thus, we proceed as follows:

[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]

In order to have 14 C atoms, 8 H atoms and 32 O atoms on both sides of the chemical equation.

Regards!

Answer:

dont buy cheap and off we went

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

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:

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:

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".

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:

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:

Anhydrous iron(III) chloride may be prepared by treating iron with chlorine:[11]

{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}

Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.

Dissolving iron ore in hydrochloric acid

{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}

Oxidation of iron(II) chloride with chlorine

{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}

Oxidation of iron(II) chloride with oxygen

{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}

Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride. Hydrated iron(III) chloride can be converted to the anhydrous form by treatment with thionyl chloride.[12] Similarly, dehydration can be effected with trimethylsilyl chloride:[13]

{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}

Anhydrous iron(III) chloride may be prepared by treating iron with chlorine.

Iron filings are small shavings of ferromagnetic material.

[tex]{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}[/tex]

Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.

Dissolving iron ore in hydrochloric acid.

Oxidation of iron(II) chloride with chlorine.

[tex]{\displaystyle {\ce {2FeCl_2_{(}aq){+~}Cl_2_{(}g)- > 2FeCl_3_{(}aq)}}}\\[/tex]

Oxidation of iron(II) chloride with oxygen.

Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride.

Hydrated iron(III) chloride can be converted to an anhydrous form by treatment with thionyl chloride. Similarly, dehydration can be affected by trimethylsilyl chloride.

[tex]{\displaystyle {\ce {FeCl_3.6H2O + 12 Me_3SiCl - > FeCl3 + 6 (Me_3Si)2O + 12 HCl}}}[/tex]

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Answer:

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

Explanation:

i think it is blood exiting the lungs

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)

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:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

Answer:

B

Explanation:

A p e x

KAnswer:

See explanation

Explanation:

It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.

Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.

Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.

Note that HCl is not a dehydrating agent.

Answer:

trigonal bipyramidal.