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Tom [10]
3 years ago
14

HELP ME I HAVE THIS REALLY IMPORTANT TEST THAT I NEED TO DO FAST!!|

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
3 0

Answer:

TRUE!

Explanation:

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Explain how to count the number of elements in a compound.
Anna35 [415]

Answer:

Simply identify what elements are in a compound

Explanation:

For example in NaCl we have sodium (Na) and Chlorine (Cl)

In order to do this you would need to recognise the symbols for a certain element: O for oxygen; N for nitrogen; H for hydrogen etc.

3 0
3 years ago
Rules for naming ionic compounds wiht transition metal
Aneli [31]

Rules for naming ionic compounds with transition metal requires the elementary name of metal followed the anion name with suffix ide.

<h3>What are ionic compounds?</h3>

Ionic compounds are made up of ions, which are charged particles that occur when an atom (or group of atoms) acquires or loses electrons. Generally cations are metals and anions are non metals in it.

Following rules will be considered during naming:

  • First determine the metal's elemental name.
  • Give the nonmetal its elemental name and the suffix -ide.
  • Use roman numerals to denote positive charge when naming metals that can have distinct oxidation states.
  • Name the polyatomic ions according to their names.

Hence rules for naming are listed above.

To know more about nomenclature of ionic compounds, visit the below link:

brainly.com/question/18672152

#SPJ4

3 0
2 years ago
If only one-fourth the amount of brown sugar in the recipe were used, how much brown sugar would be used?
melomori [17]
3/4 or 4/4 would be the answer because if you use 1/4 out of 4/4 you would either have an answer of 3/4, 4/4 or 1. The 1 because 4/4 is equal to 1. 
7 0
3 years ago
What is usually composed of two or more minerals
denis-greek [22]
Rocks and minerals are usually composed of two or more minerals
8 0
3 years ago
Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O
Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0
3 years ago
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