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Greeley [361]
3 years ago
13

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

is the nuclide symbol of X? Superscript 231 subscript 94 upper P u. Superscript 235 subscript 90 upper T h. Superscript 239 subscript 94 upper P u. Superscript 231 subscript 90 upper T h.
Chemistry
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

\rm_{90}^{231}\text{Th}

Explanation:

The unbalanced nuclear equation is

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + X

Let's write X as a nuclear symbol.

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

 92 = 2 + Z , so  Z =   92 - 2 =  90

And your nuclear equation becomes

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} +\, _{90}^{231}\text{X}

Element 90 is thorium, so  

\rm X = _{90}^{231}\text{Th}

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Pieces of copper, silver and gold are dropped into a solution of iron sulphate. The piece that will get a coating of copper is. ... Therefore, none of the three metals can displace iron from its salt solution. Hence, we observe that no reaction takes place and none of the pieces get coated.

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I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
How many molecules in each sample?<br><br> 64.7 g N2<br> 83 g CCl4<br> 19 g C6H12O6
lilavasa [31]

Answer:

  • 1.39x10²⁴ molecules N₂
  • .25x10²³ molecules CCl₄
  • 6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we <u>convert the given masses into moles</u>, using the <em>compounds' respective molar mass</em>:

  • 64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂
  • 83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄
  • 19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by <em>Avogadro's number</em>, to <u>calculate the number of molecules</u>:

  • 2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules
  • 0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules
  • 0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules
3 0
2 years ago
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Ronch [10]

Answer:

B

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