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Naya [18.7K]
3 years ago
13

Is na2S an ionic compound,covalent compound,or acid

Chemistry
1 answer:
liq [111]3 years ago
3 0

Na2S

Na is a metal, S is a non-metal, so Na2S is an ionic compound.

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Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl m
masya89 [10]

Answer:

0.535 g

Explanation:

The reaction that takes place is:

  • NaCl + AgNO₃ → AgCl + NaNO₃

First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:

  • Molarity = moles / liters
  • moles = Molarity * liters

<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L

  • moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃

Then we <u>convert AgNO₃ moles into NaCl moles</u>:

  • 0.00915 mol AgNO₃ * \frac{1molNaCl}{1molAgNO_3} = 0.00915 mol NaCl

Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:

  • 0.00915 mol NaCl * 58.44 g/mol = 0.535 g
4 0
3 years ago
PLEASE HELP!!
Luden [163]

Answer:

10 hope this helps

4 0
3 years ago
If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit
Svetllana [295]

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

4Al + 3O_{2} ⇒ 2Al_{2}O_{3}

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}

10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al

19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}

0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25

Therefore, the percent of yield is 50.03%.

4 0
3 years ago
Pls how do you test for the hydrogen ion concentration of a bottle of water.
iren [92.7K]
Using the pH=-log[ hydrogen ions], assuming the pH of water is 7 then using the concept that kw=[ OH] [H+], then the [H+]= 10 power minus 7,since kw=10 to the power minus 14
4 0
4 years ago
Boiling refers to which phase change?
Monica [59]
Boiling or also called evaporation is the conversion of liquid to gas through the application of heat. This phase change is an endothermic change and is the opposite of condensation from gas to liquid.
8 0
3 years ago
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