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2 years ago

Which contributes to the dissolution of sugar in water?

Chemistry

2 answers:

AURORKA [14]2 years ago

7 0

Option C: The dissociation of a polar covalent compound in water.

Sugar is made up of sucrose molecules containing polar covalent bonds. They have same type of oxygen-hydrogen covalent bonds that are present in water.

Since, sugar is not an ionic compound, option a and option b are incorrect. Also, due to same type of bonding in water and sugar molecules there will be no such force of attraction between them, this will opt out the option d.

Thus, dissolution of sugar in water is the dissociation of a polar covalent compound in water and option c is correct.

Veronika [31]2 years ago

3 0

I think the correct answer from the choices listed above is option d. It is the strong pull of water molecules on a polar covalent compound that contributes to the dissolution of sugar in water. Hope this answers the question.

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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne

Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0

2 years ago

Classify the following as a type of potential energy or kinetic energy (use the letters K or P)

Zanzabum

K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.

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1 year ago

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Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom

s2008m [1.1K]

D. Producers im sure of it hope im right

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2 years ago

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Name the region of the atom where protons and neutrons are located. This region is indicated by label B. Name the region of the

11Alexandr11 [23.1K]

Protons and neutrons are located at the nucleus.

Electrons are located on the outer shells, or orbitals.

Subatomic Particles are located inside of the atoms. Protons, neutrons, and electrons.

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2 years ago

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Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

nirvana33 [79]

Answer:

For 1: The correct answer is False.

For 2: The correct answer is True.

For 3: The correct answer is True.

For 4: The correct answer is False.

For 5: The correct answer is True.

Explanation:

Net ionic equation of any reaction does not include any spectator ions. If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

For 1: Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

$Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is False.

For 2: Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

$NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)$

Ionic form of the above equation follows:

$Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)$

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

For 3: Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

$Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)$

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is True.

For 4: Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

$MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the correct answer is False.

For 5: Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the correct answer is True.

3 0

1 year ago