Consider the reaction between TiO2 and C:

TiO2(s) +2C(s) → Ti(s) +CO(g)

A reaction vessel initially contains 10.0g each of the reactants. Calculate the masses of TiO2, C, Ti, and CO that will be in the reaction vessel after the reactants have reacted as much as possible

Answer:

A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.

Explanation:

Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.

Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other  releases it.

As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.

Answer:

Fronts

Explanation:

For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!

Hope this helps! Please mark as brainiest!

Answer:

61.5%.

Explanation:

We usually define percentage yield in chemistry as follows;

Percentage yield= actual yield/theoretical yield ×100

In this case we have;

Actual yield of the guaifenesin 400mg

Theoretical yield of the guaifenesin 650 mg

Hence;

Percentage yield of the guaifenesin =

400/650 ×100 = 61.5%

Hence the theoretical yield of the guaifenesin is 61.5%.

Answer:

i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances

Explanation:

Answer:

None

Explanation:

Answer on Edge 2022

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

Answer:

THE NEW FREEZING POINT IS -4.196 °C

Explanation:

ΔTf = 1 Kf m

molarity of MgCl2:

Molar mass = (24 + 35.5 *2) g/mol

molar mass = 95 g/mol

7.15 g of MgCl2 in 100 g of water

7.15 g = 100 g

(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3

= 0.715 g /dm3

Molarity in mol/dm3 = molarity in g/dm3 / molar mass

= 0.715 g /dm3 / 95 g/mol

m = 0.00752 mol/ dm3

So therefore:

ΔTf = i Kf m

1 = 3 (1 Mg and 2 Cl)

Kf = 1.86 °C/m

M = 0.752 moles

So we have:

ΔTf = 3 * 1.86 * 0.752

ΔTf = 4.196 °C

The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C

Answer:

2.1 atm

Explanation:

Before we get the total pressure, we have to ensure all the gases have the same pressure unit.

Nitrogen gas = 0.75 atm

Helium = 490mmHg

To convert mmHg to atm;

760 mmHg = 1 atm

490 = x

x = 460 / 760 = 0.645 atm

Neon = 520 torr

Converting torr to atm;

760 torr = 1 atm

520 torr = x

x = 520 / 760 = 0.684 atm

The total pressure is then given as;

0.75 + 0.684 + 0.645 = 2.1 atm

Answer: single replacement reaction

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.

A general single displacement reaction can be represented as :

[tex]X+YZ\rightarrow XZ+Y[/tex]

The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]

When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

Answer:

[tex]\large \boxed{\text{0.195 mol/L}}[/tex]

Explanation:

(a) Balanced equation

2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O  

(b) Moles of Ca(OH)₂

[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]

(c) Moles of HNO₃

The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂

[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]

(d) Molar concentration of HNO₃

[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]

Answer:

Explanation:

Phosphoric acid H₃PO₄ is produced from reaction of water and tetraphosphorus decoxide  P₄O₁₀ as follows

P₄O₁₀ + 6 H₂O = 4 H₃PO₄

In this reaction 6 molecules of H₂O and one molecule of  phosphorus compound P₄O₁₀ is needed to produce phosphoric acid , ie the conversion factor of water to acid is 6 / 1

This ratio is called mole ratio of water to H₃PO₄.

So the required ratio is 6 : 1 .

Answer:

Intermediate.

Explanation:

Hello,

In this case, we can rewrite the steps as:

[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]

Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).

Best regards.

Answer:

[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]

The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

        Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]

           [tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]

So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

Find out the more information about the concentration here:

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Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

Answer:

The answer is " 10.39"

Explanation:

Calculating acid moles:

[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]

Calculating NaOH moles:

[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]

calculating excess in OH-  Moles:

[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]

calculating total volume:

[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]

[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]

           [tex]=0.00025 M[/tex]

[tex]pOH = - \log 0.00025[/tex]

        = 3.6

[tex]pH = 14 - pOH[/tex]

      = 10.39

Answer:

Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.  

Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.  

Answer:

pH = 5.24

Explanation:

Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.

Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:

[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M

[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.76M] / [0.24M]

Answer: -1,030 kJ

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex]     .....(1)

Putting values in equation 1, we get:

[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]

The balanced chemical reaction is:

[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]  [tex]\Delta H=-726kJ[/tex]

Given :

Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J

Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts =  [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]

Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]

HERE IS THE COMPLETE QUESTION

what is the calculated value of the cell potential at 298 k for an electrochemical cell with the following reaction, when the Cl2 pressure is 1.31 atm, the cl- concentration is at  1.28M, and the Ni2+ concentration is 1.24M

[tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

Answer:

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

Explanation:

From the given question;

We can see that Nickel is oxidized and Chlorine is reduced. We also known that Oxidation occurs at the anode while reduction occurs at the cathode.

SO the Oxidation and the reduction reaction can be expressed as shown below.

Oxidation :      [tex]Ni_{s} \to Ni ^{2+}_{aq} + 2e^- \ \ \ \ E^0 = -0.26[/tex]

Reduction :      [tex]Cl_{2(g)} + 2e^- \to 2Cl^- _{(aq)} \ \ \ \ E^0} = 1.36[/tex]

                        [tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

[tex]E^0_{cell} = E^0_{cathode}-E^0_{anode}[/tex]

[tex]E^0_{cell} = E^0_{Cl_2/Cl^-}-E^0_{Ni^{2+}/Ni}[/tex]

[tex]E^0_{cell} = +1.36 - (-0.26)[/tex]

[tex]E^0_{cell} = +1.62 \ V[/tex]

By applying Nernst Equation; we have :

[tex]E_{cell} = E^0_{cell}- (\dfrac{0.0591}{n}) \ log Q[/tex]

where

n= 2 moles

[tex]Q= \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[1.28]^2[1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384][1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384*1.24]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[2.031616]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085*10^{4})[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085)+ log \ 10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 0.1905 + 4[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 4.1905[/tex]

[tex]E_{cell} = +1.62 \ V}- 0.1238[/tex]

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

The given question is incomplete, the complete question is:

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer:

The correct answer is option D, that is, -349 kJ/mol.

Explanation:

Based on the given information, the reaction is:  

NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.  

Therefore, there is a need to reverse the reaction and change the sign on ΔG.  

Now the reaction will become,  

Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).  

Answer:

[tex]Q=4154J[/tex]

Explanation:

Hello,

In this case, the involved heat in this heating process is considered to be computed via:

[tex]Q=nCp\Delta T[/tex]

Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:

[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]

Regards.

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

Answer:

Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.

Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.

The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.

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Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.187

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

Where [] can be understood as moles of each specie.

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

Initial moles:

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate

Moles after reaction:

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

Answer:

Thin-layer chromatography

Explanation:

Here in this question, we are concerned with giving the best technique that could be use for monitoring the progress of the reaction that involves the oxidation of alcohols.

The most useful technique here for monitoring the progress of the reaction would be the thin layer chromatography.

Firstly, we should know the reason why we are using a chromatographic technique. This is intuitive. The reason is that the oxidizing agent being used is a colored substance. So basically, the progress of the oxidation reaction would solely rest on the fact that we have a decrease in color saturation of the said oxidizing agent.

This is best done using the thin layer chromatography because we have a mobile phase that moves quickly over the stationary phase and asides this, it moves evenly.

This makes the fact that color changes can be quickly and easily identified and we can tell if the oxidation reaction has gone to completion or not

Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.

Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.

It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.

Answer:

C.

Explanation:

Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.

Answer:

(a) The empirical formula of the compound is

m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).

(b) The grams of O2 that were used in the reaction is 1.146 g

(c) The amount of O2 that would have been required for complete combustion is 1.401 g.

Explanation:

a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)

(b) Using law of conservation of mass from above

m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)

m(O2) = 0.446 + 0.700 + 0.430 - 0.430

m(O2) = 1.146 g

The grams of O2 that were used in the reaction is 1.146 g

(c) for complete combustion, we need to oxidized CO to CO2

Then, 2CO +O2 = 2CO2

m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}

m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g

Note; Molar mass of O2 = 32, CO = 28

m(total)(O2) = m(O2) + m(add)(O2)

m(total)(O2) = 1.146 + 0.255 = 1.401 g

The amount of that grams would have been required for complete combustion is 1.401 g.

Note (add) and (total) were used subscript to "m"

Answer:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]

I hope this explanation is clear and explanatory.