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2 years ago

How many electrons are in a atom of zicronium

Chemistry

2 answers:

Karo-lina-s [1.5K]2 years ago

5 0

Answer:

Name Zirconium

Atomic Mass 91.224 atomic mass units

Number of Protons 40

Number of Neutrons 51

Number of Electrons 40

Explanation:

MrRissso [65]2 years ago

3 0

Answer:

There are 40 electrons in one atom of Zirconium.

Explanation:

Note: The word is not zicronium, it is Zirconium.

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Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

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

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

5 0

2 years ago

Give one example of fast erosion and one example of slow erosion. Use complete sentences.

topjm [15]

An example of erosion is the Grand Canyon, which was worn away over time by the Colorado river.

3 0

2 years ago

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Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

#SPJ4

8 0

1 year ago

An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the

sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

$P_i$ = 100 N·m

$T_i$ = 327 K

$P_f$ = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln( $P_f$ / $P_i$ ) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈ -0.52571 kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln( $P_f$ / $P_i$ ) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = -7.5 kJ

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = -5.00574 kJ

6 0

2 years ago

State law of conservation of mass and prove the law for below given chemical reaction

lawyer [7]

The law of conservation of mass say that, in a chemical reaction, the mass of the reagents will always be equal to the mass of the products. This is shown in the reaction given below.

Mass of the reagent: 100 g.
Mass of the products: 56 + 44 = 100 g.

8 0

1 year ago

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