if excess nitrogen gas reacts with 600 cm³ of hydrogen gas at room conditions , calculate the maximum volume of ammonia produced from the reaction? the chemical question is N2 + 3H2 = 2NH3​

Answer: The given amount of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 0.430 moles of Cd, 0.860 moles of N and [tex]2.59\times 10^{23}[/tex] molecules of water.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of [tex]Cd(NO_3)_2.4H_2O[/tex] = 132.4 g

Molar mass of [tex]Cd(NO_3)_2.4H_2O[/tex] = 308.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of }Cd(NO_3)_2.4H_2O=\frac{132.4g}{308.5g/mol}=0.430 mol[/tex]

1 mole of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 1 mole of Cd, 2 moles of nitrogen atom (N), 10 moles of oxygen atom (O) and 8 moles of hydrogen atom (H).

So, 0.430 moles of [tex]Cd(NO_3)_2.4H_2O[/tex] will contain = [tex](1\times 0.430) = 0.430mol[/tex] of Cd and [tex](2\times 0.430)=0.86mol[/tex] of N

According to the mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules

So, 0.430 moles of [tex]Cd(NO_3)_2.4H_2O[/tex] will contain = [tex]\frac{6.022\times 10^{23}}{1mol}\times 0.430mol=2.59\times 10^{23}[/tex] number of water molecules.

Hence, the given amount of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 0.430 moles of Cd, 0.860 moles of N and [tex]2.59\times 10^{23}[/tex] molecules of water.

Explanation:

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Explanation:

here's the answer to your question

double arrow just mean that it's a reversible process, and the reaction can go back and forth.

Answer:

nitrogen is used in the production of  1) fertilisers  2) nitric acid  3) nylon  4) dyes  and  5) explosives

please give me brainliest if you can :))

Answer:

the reactants are carbon dioxide ,iron, oxygen

Answer:

C2H4 Ethylene

Explanation:

Hope it helps! :)

Answer:

sodium chloride

hope that helped :)

Answer:

Chlorine is a greenish yellow gas at room temperature and atmospheric pressure. ... yielding chlorine water, and from this solution a solid hydrate of ideal ...

Answer:

Nitrogen oxides are used in the production of nitric acid, lacquers, dyes, and other chemicals. Nitrogen oxides are used in rocket fuels, in the nitrification of organic chemicals, and in the manufacture of explosives.

Explanation:

Answer:

oxygen molecule has two oxygen atom . Each O atom share 2 electrons to form two covalent bonds out of which one is sigma bond and other is pi bond . sigma bond is formed by axial overlap 2p atomic orbitals of oxygen and pi bond is formed of lateral overlap of 2p atomic orbitals of oxygen .

Answer:

Acid is a molecule capable of donating hydrogen ion and they form aqueous solutions with a sour taste while base is a substance that accepts proton from proton donor and in aqueous solution, they have an astringent or bitter taste. Moreover, a good example for base is sodium hydrogen carbonate as baking soda or baking powder and for acid, the most common example is the acetic acid or vinegar.

Answer:

Well, for starters, bases mostly consist of soaps and cleaning products, acids consist of vinegar,lemon,and stuff like that. Most acids on the basis of touch can burn your skin or be very sour, unlike bases which are more soapy and have a cleaner touch.A good example of an acid is vinegar, which can be used for cleaning but can also be used for making foods, on the other hand. A good example of a base is dish soap. It is used for cleaning.

Explanation:

i just did the exam and got a 100

Answer:

Create a new account as the thing you're experiencing now is due to the fact that Brainly had blocked your account.

Answer:

A 1.732 g sample of iron is heated in air. It oxidizes to form a product with a mass of 2.205 g. What is the empirical formula of the product? FeO 4. Caffeine is made of 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen by mass.

Explanation:

The empirical formula of the product formed by the reaction of iron and oxygen is FeO.

Let's consider the following generic chemical equation:

Fe + O₂ ⇄ FexOy

According to the law of conservation of mass, the sum of the masses of the reactants is equal to the sum of the masses of the products. The mass of O₂ that reacted, and remains in the oxide is:

[tex]mFe + mO_2 = mFexOy\\mO_2 = mFexOy - mFe = 2.205 g - 1.732 g = 0.473 g[/tex]

The oxide has 1.732 g of Fe and 0.473 g of oxygen and a total mass of 2.205 g. To determine the empirical formula, we need to calculate the percent composition.

[tex]\%Fe = \frac{mFe}{mFexOy} \times 100 \% = \frac{1.732 g}{2.205 g} \times 100 \% = 78.55\%[/tex]

[tex]\%O = \frac{mO}{mFexOy} \times 100 \% = \frac{0.473 g}{2.205 g} \times 100 \% = 21.45\%[/tex]

Now, we will divide each percentage by the atomic mass of the element.

[tex]Fe: 78.55/55.85 = 1.406\\O: 21.45/16.00 = 1.340[/tex]

Finally, we divide both numbers by the smallest one.

[tex]Fe: 1.406/1.340 \approx 1 \\O: 1.340/1.340 = 1[/tex]

The empirical formula of the product is FeO.

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Answer:

69

Explanation:

I thought that I answer is correct

Answer:

The answer is 69

Explanation:

Because if you add one its 66, and so on, hope this helps! but if you add five it will be 69, if you need help with easy equations let me know

Answer:

62.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.

Thus, 62.5 moles of O₂ is needed for the reaction.

Answer:

Answer: Atoms of chlorine (CI) and magnesium (Mg)

Answer: Atoms of chlorine (CI) and magnesium (Mg)is your answer

Answer:

the answer is 47.9 and ik because I just had that question

The empirical formula of the compound is O₄Pb₃.

An Empirical system is the chemical system of a compound that offers the proportions of the elements gifted within the compound however not the real numbers or arrangement of atoms. This would be the lowest complete variety ratio of the elements within the compound.

Amount of lead (Pb) = 90.6%

⇒and amount of oxygen = 9.4%

taking the whole number ratio

o = 4

Pb = 3

∴ ⇒O: Pb=4:3

O4Pb3 answer.

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Answer: broooo what is this

Explanation:

The heat of hydration is defined as the heat absorbed or evolved when one mole of a substance undergoes hydration. The heat of solution is the enthalpy change associated with he dissolution of a solute. The lattice energy is the heat. Lattice energy is the energy released when the components of the lattice are brought together from infinity.

Hence the heat of solution for LiF → Li^+ + F- is -1966  kJ/mol

Given that;

Heat of hydration = ΔH solution – ΔH lattice energy

Where,

ΔH solution  = Heat of the solution

ΔH lattice energy = Lattice energy of the solution

The heat of solution or enthalpy of dissolution is defined as the enthalpy change associated with the dissolution of a solute in a solvent

From the formula above;

ΔH solution = Heat of hydration + ΔH lattice energy

Heat of hydration = [(-431) + (-499)] = -930 kJ/mol

ΔH solution = (-930) + (-1,036) = -1966  kJ/mol

Hence the heat of solution for LiF → Li^+ + F- is -1966  kJ/mol

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Explanation:

it is that when an object is moving 25 m per sec to calculate its time we can simply divide 1.5 by 25 and multiply by 1

Answer:

a. 1 x 10^-35

Explanation:

The correct compound given is: [tex]\mathsf{Cu_3(AsO_4)}_2[/tex]

To predict the approximate Ksp value of the given compound, we will need to express the oxidation-reduction half-reaction of the compound and its dissociation, then, we will use the Nernst equation to determine the approximate Ksp value.

To start with the reduction half-reaction:

[tex]\mathsf{Cu_3(AsO_4)_{2}(s) + 6e^- \to 2As O_{4}^{3-}_{(aq)}+3Cu(s) }[/tex]

The oxidation half-reaction is:

[tex]\mathsf{3Cu(s) \to 3CU^{2+}_{(aq)} + 6e^-}[/tex]

The overall cell reaction now is:

[tex]\mathsf{Cu_3(AsO_4)_{2}(s) \to 3Cu^+ (aq) + 2As O_{4}^{3-}_{(aq)} }[/tex]

From the reduction half-reduction, the number of moles of electrons (n) transferred is 6 moles.

By applying the Nernst equation:

[tex]\mathsf{E_{cell} = E^0_{cell} -\dfrac{0.0591V}{n}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }[/tex]

At standard conditions;

The standard cell potential [tex]\mathsf{E^0_{cell} = -0.342 \ V}[/tex]

and [tex]\mathsf{E_{cell} = 0 \ V}[/tex] since it is at equilibrium.

∴

[tex]\mathsf{0 = -0.342 -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 } \\ \\ \\ \mathsf{0.342 = -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }[/tex]

[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = \dfrac{-(0.342)*6}{0.0591 }}[/tex]

[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = -34.7}[/tex]

[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 \simeq -35}[/tex]

[tex]\mathsf{[Cu^{2+}]^3[AsO_4^{3-}]^2 = 10^{-35}}[/tex]

[tex]\mathbf{K_{sp} = [Cu^{2+}]^3[AsO_4^{3-}]^2 = 1\times 10^{-35}}[/tex]

Answer:

See explanation

Explanation:

Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.

In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.

Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.

Ca^2+ ion  is smaller than Mg^2+ ion hence it has a greater heat of formation or lattice energy.

When a crystal lattice is formed, the energy that is released if the component ions of the compound are brought together from infinity is called the lattice energy. It is the energy released when a crystal lattice is formed from its component ions.

The question lets us know that the heat released by the compounds depends on the energy released upon formation of the compound. Hence, the higher the energy released upon formation, the higher the magnitude of heat released upon dissolution of the compound.

Recall that lattice energy depends on the size of the ions. Thus, the smaller the ions, the higher the lattice energy.

Ca^2+ is smaller than Mg^2+ hence more energy is given off when CaCl2  is formed than when MgCl2  is formed.

As stated above, the greater the lattice energy, the greater the heat released when the lattice dissolves and the higher the rise in temperature.

Putting all these together, Sample A must be CaCl2 while sample B must be MgCl2.

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Answer:  1)  temperature, 2) add a catalyst, and 3) concentrations

Explanation:  A rise in temperature adds kinetic energy to the reactants, leading to more frequent, and energetic collisions, which will often accelerate the reaqction.  A catalyst is an agent that helps the reaction move forward more quickly by offering a substrate that aids in orienting the reactants for more efficient reaction.  A higher concentration increases the chances of collisons that result in products.

Answer:

It because ,sodium is a metal and chlorine is a non metal

Force = The external energy that changes or tends to change the state of any body or object is called force.

Speed = The rate of distance is called speed.

Answer:

Force is a push or pull which changes or tend to change the position of a body.

The rate of change its position with time or magnitude is called speed.

Answer:

Pb2+ (aq) + CrO42-(aq) --> PbCrO4(s) - A precipitation reaction

2 Mg(s) + O2(g) -> 2 MgO(s) - An oxidation-reduction reaction that is also a synthesis reaction

Ag^+ (aq) + 2NH3(aq) --> [Ag(NH3)2]^+ (aq) - A Lewis acid-base reaction the produces a coordination complex

Explanation:

A precipitation reaction is one in which two aqueous reactants yields an insoluble product called a precipitate as shown in reaction 1 above.

In the second reaction Mg is oxidized from zero to + 2 while oxygen was reduced from zero to -2 as the MgO is formed hence the reaction is an oxidation-reduction reaction that is also a synthesis reaction.

In the third reaction, Ag^+ a Lewis acid reacts with NH3, a lewis base to yield the complex [Ag(NH3)2]^+.

Answer:

[tex]\boxed {\boxed {\sf 2.36 \ mol \ SO_2}}[/tex]

Explanation:

We are asked to convert grams to moles. We will use the molar mass and dimensional analysis to perform this conversion.

The molar mass is the mass of 1 mole of a substance. These values are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead.

We are given a mass of sulfur dioxide (SO₂). Look up the molar masses of the individual elements.

Notice that the formula of the compound contains a subscript. The subscript after O means there are 2 moles of oxygen in 1 mole of sulfur dioxide. We must multiply oxygen's molar mass before adding sulfur's.

Now we will use dimensional analysis to convert grams to moles. From the molar mass, we know there are 64.068 grams of sulfur dioxide per mole, so we can set up a ratio.

[tex]\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}[/tex]

We are converting 151 grams to moles, so we multiply by this value.

[tex]151 \ g \ SO_2 *\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}[/tex]

Flip the ratio so the units of grams of sulfur dioxide cancel.

[tex]151 \ g \ SO_2 *\frac {1 \ mol \ SO_2}{64.068 \ g \ SO_2}[/tex]

[tex]151 *\frac {1 \ mol \ SO_2}{64.068 }[/tex]

[tex]\frac {151}{64.068 } \ mol \ SO_2[/tex]

[tex]2.356870825 \ mol \ SO_2[/tex]

The original measurement of grams (151) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 6 in the thousandth place tells us to round the 5 in the hundredth up to a 6.

[tex]2.36 \ mol \ SO_2[/tex]

151 grams of sulfur dioxide is approximately 2.36 moles of sulfur dioxide.

Explanation:

it helps to make juices.

It helps to make concentrated acid into dilute acid.

Answer:

Explanation:

Here are a few more examples:

1. Smoke and fog (Smog)

2. Dirt and water (Mud)