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3 years ago

Classify which compounds will dissolve in water and which ones will not dissolve in water.

Chemistry

1 answer:

vredina [299]3 years ago

6 0

Answer:

Dissolves in water: Acetone, 1-propanol, Methanol

Do not dissolve in water: Hexane, Decane

Explanation:

From the principle that like dissolves like,polar substances will dissolve in polar solutions, while non-polar substances will dissolve in non-polar substances.

Water is a polar solution, therefore, polar substances will dissolve in it.From the given options:

Acetone is a polar molecule, therefore, it will dissolve in water.

1-propanol is a polar molecule, therefore, it will dissolve in water.

Hexane is a non-polar molecule, therefore, it will not dissolve in water.

Methanol is a polar molecule, therefore, it will dissolve in water.

Decane is a non-polar molecule, therefore, it will not dissolve in water

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Which element can join with other elements to form covalent bonds? beryllium (Be) carbon (C) cesium (Cs) iron (Fe)

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Answer:

Carbon (C)

Explanation:

Carbon is the only nonmetal there and covalent bonds happen between two nonmetal atoms

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3 years ago

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In the reaction fecl2 + 2naoh fe(oh)2(s) + 2nacl, if 6 moles of fecl2 are added to 6 moles of naoh, how many moles of fecl2 woul

pychu [463]

The balanced chemical reaction would be

fecl2 + 2naoh = fe(oh)2(s) + 2nacl

Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:

6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2

Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.

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3 years ago

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A little aluminum boat (mass of 14.50

Valentin [98]

When the volume is 450 cm3 so the boat will displace 450 g.

so we have 450 g - 14.5g = 435.5 left

so to get how many pennies can be added to the boat before it sinks can be determined by:

divided the mass for the one penny:

no.of Penny = 435.5 / 2.5g

= 174.2

∴ the boat will float with 174 pennies and will start to sink with 175

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3 years ago

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Kindly checkout the attached pic. It contains all your required questions.

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3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

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3 years ago