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3 years ago

How does induced transmutation produce trans uranium elements ?

1 answer:

5 0

Answer: Artificial transmutation may occur in machinery that has enough energy to cause changes in the nuclear structure of the elements. This releases, on average, 3 neutrons and a large amount of energy. The released neutrons then cause fission of other uranium atoms, until all of the available uranium is exhausted.
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5 0

2 years ago

Barium nitrate has the formula Ba( NO3)2. Which statement is true about barium nitrate?

OLga [1]

I found these four statements for that question:

Each molecule contains four different elements.

Each molecule contains three atoms.

Each molecule contains seven different bonds.

Each molecule contains six oxygen atoms.

The last one is true. Each molecule contains six oxygen atoms.
The number to the right of O and of (NO3) ares subscripts.

The chemical formula uses subscripts to indicate the number of atoms.

The subscript 2 in (NO3)2 means that there are two NO3 radicals.

And the subscript 3 to the right of O means that each NO3 radical has three atoms of O.

Then, the number of atoms of O is 2 * 3 = 6.

So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.

From that molecule you can also tell:

- Each molecule contains one atom of barium

- Each molecule contains two atoms of nitrogen

- Each molecule contains two NO3 radicals

4 0

3 years ago

2Mg + O2 → 2MgO If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

LekaFEV [45]

<h3>Answer:</h3>

9.6724 g MgO

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

Step 2: Identify Conversions

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})$
Multiply/Divide: $\displaystyle 9.67241 \ g \ MgO$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

9.67241 g MgO ≈ 9.6724 g MgO

5 0

3 years ago

Read 2 more answers

Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O

Feliz [49]

Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺

3. Separate the skeleton equation into two half-reactions.

PbO₂ ⟶ Pb²⁺

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

PbO₂ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}$

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

7 0

3 years ago

Ayuda dame un ejemplo de fusión!!!

Rashid [163]

?? What do you mean

6 0

3 years ago