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TiliK225 [7]
3 years ago
7

What element is denoted by 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1

Chemistry
2 answers:
Anestetic [448]3 years ago
7 0

Element 39....Yttrium

alexgriva [62]3 years ago
4 0

The answer is Zr (Zirconium)

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Rainforests are not generally considered a source of _______.
schepotkina [342]

Rainforests are not generally considered a source of none of the above choices. You can get food from Rainforests, and also spices and medicines. In fact, Rainforests have a variety of plants that have medicinal properties. When you know how to find the right plant in the rainforests, you can access almost everything.

<h2>Hence, your answer is D. None of the above</h2>

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Answered by: FieryAnswererGT

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8 0
2 years ago
How many valence electrons are in Lead(IV)
Ksenya-84 [330]

Answer:

It has 4 valence electrons

7 0
3 years ago
Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate
Bingel [31]

Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • We can balance the equation by applying the conservation of mass to the equation.
  • The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
  • This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
  • We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass. </em>

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

<u><em>Using cross multiplication: </em></u>

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>

3 0
3 years ago
21+ Cl2 → 12+201 -
Alina [70]

Answer:

\rm 2\; I^{-} + Cl_2 \to I_2 + 2 \; Cl^{-}.

Start color: yellowish-green.

End color: dark purple.

Assumption: no other ion in the solution is colored.

Explanation:

In this reaction, chlorine gas \rm Cl_2 oxidizes iodine ions \rm I^{-} to elemental iodide \rm I_2. At the same time, the chlorine atoms are converted to chloride ions \rm Cl^{-}.

Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.

Among the reactants of this reaction, \rm I^{-} is colorless. If there's no other colored ion, only the yellowish-green hue of \rm Cl_2 would be visible. Hence the initial color of the reaction would be the yellowish-green color of \rm Cl_2.

Similarly, among the products of this reaction, \rm Cl^{-} is colorless. If there's no other colored ion, only the dark purple hue of \rm I_2 would be visible. Hence the initial color of the reaction would be the dark purple color of \rm I_2.

5 0
3 years ago
You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne
fenix001 [56]

Answer:

P_2=1.90atm

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}

Whereas both n and R are cancelled out as they don't change, we obtain:

\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}

Thus, by solving for the final pressure, we obtain:

\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm

Best regards!

7 0
2 years ago
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