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Romashka-Z-Leto [24]
1 year ago
12

Write balanced nuclear equations for the following:(b) Alpha decay of polonium-218

Chemistry
1 answer:
Digiron [165]1 year ago
7 0

The balanced nuclear equations for the following:(b) Formation of polonium-218 through α decay is (210,83)Bi -> (4,2)alpha + (210,81)Po

<h3>What is balanced nuclear equation?</h3>

A nuclear reaction is generally expressed by a nuclear equation, which has the general form, where T is the target nucleus, B is the bombarding particle, R is the residual product nucleus, and E is the ejected particle, and Ai and Zi (where I = 1, 2, 3, 4) are the mass number and atomic number, respectively. Finding a well balanced equation is critical for understanding nuclear reactions. Balanced nuclear equations provide excellent information about the energy released in nuclear reactions. Balancing the nuclear equation requires equating the total atomic number as well as the total mass number before and after the reaction using the rules of atomic number and mass number conservation in a nuclear reaction.

To learn more about nuclear equations visit:

brainly.com/question/12221598

#SPJ4

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1. john needs to create a buffered solution at a ph of 3.5 for his biomedical laboratory
Lunna [17]

Answer:

Use a ratio of 0.44 mol lactate to 1 mol of lactic acid  

Explanation:

John could prepare a lactate buffer.

He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.

\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}

He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.

For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.

6 0
3 years ago
If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)
Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln}    *100\%\\\\w/w\%=3.15\%

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0
3 years ago
Consider the redox reaction below.
vovangra [49]

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

4 0
3 years ago
What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma
Montano1993 [528]

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

                     1 g of H ----------------- 1 mol

                   3,25 g of H -------------  x

                     x = (3.25 x 1) / 1

                     x = 3.25 moles

                    12 g of C ---------------- 1 mol

                     19.36 g of C ----------  x

                     x = (19.36 x 1) / 12

                     x = 1.61 moles

                     16g of O --------------- 1 mol

                     77.39 g of O ---------  x

                      x = (77.39 x 1)/16

                      x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

                        C₂HO₃

4 0
3 years ago
se tienen 10 litros de nitrogeno gaseoso bajo una presion de 560 mm Hg. Determina el volumen del nitrogeno a una presion de 760
Dominik [7]

Answer:

working business city philadelphian

3 0
2 years ago
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