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3 years ago

What is the mass of phosphorus that contains twice the number of atoms found in 14

Chemistry

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

15.5 gm

Explanation:

What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron?

[Relative atomic mass : P = 31; Fe = 56]

14 gm Fe = 14gm/ 56 gm/mole = 14 mole gm/56gm = 14/56 mole

0.25 moles

2 X 0.25 = 0.5 moles

1 mole P = 31 gm

0.5 moles P =31/2 =15.5 gm

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Equations can be balanced by using the half-reaction method. Which step should be completed first when using this method?

Tamiku [17]

Answer:

Separate the reaction into half equations

Explanation:

When we want to balance redox equations, we must take cognizance of the fact that a specie was oxidized and another specie was reduced.

Hence we must identify the specie that was oxidized and the one that was reduced and then break up the whole redox reaction into oxidation and reduction half equations.

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What changes would occur at a molecular level if a liquid is placed in cool conditions?

Dafna1 [17]

Answer: Option (C) is the correct answer.

Explanation:

Movement of particles in a substance is responsible for change in state of the substance or matter.

This means that more is the motion of particles more will be their kinetic energy.

Also, kinetic energy is directly proportional to temperature.

K.E = $\frac{3}{2}kT$

So, less is the temperature of an object or substance less will be be the motion of its particles. Therefore, molecules will come closer to each other and state of substance will change from liquid to solid.

Thus, we can conclude that the motion of the molecules would decrease at a molecular level if a liquid is placed in cool conditions.

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3 years ago

Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

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When air bags inflate, nitrogen gas is formed from sodium azide, along with solid sodium. what reaction category is this?

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The answer is
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3 years ago

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Digiron [165]

Species shown in bold are precipitates.

Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃

Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃

Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃

Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃

A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.

Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.

Compare the first and last row:

Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.

As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.

Compare the second and third row:

Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.

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3 years ago