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amm1812
2 years ago
10

I do not understand this question if someone could help please

Chemistry
1 answer:
zzz [600]2 years ago
4 0

\bold{\huge{\blue {\underline{ Answer}}}}

  1. <em>Potassium </em><em>=</em><em> </em><em>K3PO4</em><em> </em><em>,</em><em> </em><em>K2O</em><em>, </em><em>KF</em><em>, </em><em>KBr</em><em>, </em><em>KNO2</em><em> </em><em>,</em><em> </em><em>K2SO4</em><em> </em>
  2. <em>Sodium </em><em>=</em><em> </em><em>Na3PO4</em><em> </em><em>,</em><em> </em><em>Na2O</em><em>, </em><em>NaF</em><em>, </em><em>NaBr</em><em>, </em><em>NaNO</em><em>3</em><em> </em><em> </em><em>,</em><em> </em><em>Na2SO4</em>
  3. <em>Magnesium </em><em>=</em><em> </em><em> </em><em>Mg3</em><em>(</em><em>P</em><em>O</em><em>4</em><em>)</em><em>2</em><em> </em><em>,</em><em> </em><em>MgO</em><em>, </em><em>MgF2</em><em> </em><em>,</em><em> </em><em>MgBr2</em><em> </em><em>,</em><em> </em><em>Mg3NO2</em><em> </em><em>,</em><em> </em><em>MgSO4</em><em> </em>
  4. <em>Aluminium</em><em> </em><em>=</em><em> </em><em>AlPO4</em><em> </em><em>,</em><em> </em><em>Al2O3</em><em> </em><em>,</em><em> </em><em>AlF3</em><em> </em><em>,</em><em> </em><em>AlBr3</em><em> </em><em>,</em><em> </em><em>AlNO2</em><em> </em><em>,</em><em> </em><em>Al2</em><em>(</em><em>S</em><em>O</em><em>4</em><em>)</em><em>3</em>
  5. <em>Iron(</em><em>|</em><em>|</em><em>|</em><em>)</em><em> </em><em>=</em><em> </em><em>FePO4</em><em> </em><em>,</em><em> </em><em> </em><em>Fe2O3</em><em> </em><em>,</em><em> </em><em>FeF3</em><em> </em><em>,</em><em> </em><em>FeBr3</em><em> </em><em>,</em><em> </em><em>F</em><em>e</em><em>N</em><em>O</em><em>2</em><em> </em><em>,</em><em> </em><em>Fe2</em><em>(</em><em>S</em><em>O</em><em>4</em><em>)</em><em>3</em><em> </em>
  6. <em>Copper(</em><em>|</em><em>|</em><em>)</em><em> </em><em>=</em><em> </em><em>Cu3</em><em>(</em><em>P</em><em>O</em><em>4</em><em>)</em><em>2</em><em> </em><em>,</em><em> </em><em>CuO</em><em>, </em><em>CuF2</em><em> </em><em>,</em><em> </em><em>CuBr2</em><em> </em><em>,</em><em> </em><em>Cu3NO2</em><em> </em><em>,</em><em> </em><em>CuSO4</em>
  7. <em>Barium </em><em>=</em><em> </em><em>Ba3</em><em>(</em><em>P</em><em>O</em><em>4</em><em>)</em><em>2</em><em> </em><em>,</em><em> </em><em>BaO</em><em>, </em><em> </em><em>BaF2</em><em> </em><em>,</em><em> </em><em>BaBr2</em><em> </em><em>,</em><em> </em><em>Ba3NO2</em><em> </em><em>,</em><em> </em><em>BaSO4</em><em> </em><em>.</em>
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4 years ago
What is the formula and name of the compound produced at the end of the virtual lab, and what caused it to have a different appe
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Answer: SnO_2 , NO_2 and H_2O are formed at the end of the reaction. They are named as tin (IV) oxide or stannic oxide, nitrogen dioxide and water respectively.

Explanation: Reaction of tin and nitric acid is given as:

Sn(s)+4HNO_3(aq.)\rightarrow SnO_2(s)+4NO_2(g)+2H_2O(l)

Three products are formed at the end of the reaction which are:

  • SnO_2 which is termed as stannic oxide or Tin (IV) oxide. This is a white colored solid.
  • NO_2 which is termed as nitrogen dioxide. These are brown colored fumes.
  • H_2O which is termed as water.

At the starting tin was a silvery-white colored solid and after the reaction, it changed the color to milky-white. This change in color is due to the chemical reaction happening between tin and nitric acid.

Release of brown fumes are also an indication that a chemical reaction has taken place.

6 0
3 years ago
How many milliliters of 0.121 M HCl are needed to neutralize 44.3 mL 0.274 M Ba(OH)2? Write the equation and don't forget to bal
Sergeeva-Olga [200]

Answer:

100 mL is the volume of HCl needed to neutralize the 44.3 of Ba(OH)₂

Explanation:

This is the chemical equation:

2HCl + Ba(OH)₂  →  BaCl₂ + 2H₂O

Formula for neutralization is:

Molarity of acid . Volume of acid = Molarity of base . Volume of base

0.121 . Volume of acid = 0.274 . 44.3

Volume of acid = (0.274 . 44.3) / 0.121 → 100

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Answer:

D

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4 0
4 years ago
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