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3 years ago

Net ionic equation of HF(aq)+RbOH(aq)=H2O(l)+RbF(aq)

Chemistry

1 answer:

Tamiku [17]3 years ago

4 0

net ionic equation

H⁺(aq)+OH⁻(aq)⇒H₂O(l)

<h3>Further explanation </h3>

The electrolyte in the solution produces ions.

The equation of a chemical reaction can be expressed in the equation of the ions

For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule

In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

So only the dissolved compound is ionized ((expressed in symbol aq)

Reaction

HF(aq)+RbOH(aq)⇒H₂O(l)+RbF(aq)

ionic equation

H⁺(aq)+F⁻(aq)+Rb⁺(aq)+OH⁻(aq)⇒H₂O(l)+Rb⁺(aq)+F⁻(aq)

net ionic equation

H⁺(aq)+OH⁻(aq)⇒H₂O(l)

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How many moles of methanol must be added to 4.50 kg of water to lower its freezing point to -11.0 ∘c? for each mole of solute, t

DochEvi [55]

<u>Given:</u>

Mass of solvent water = 4.50 kg

Freezing point of the solution = -11 C

Freezing point depression constant = 1.86 C/m

<u>To determine:</u>

Moles of methanol to be added

<u>Explanation:</u>

The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:

ΔTf = kf*m

where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution

ΔTf = 0 C - (-11.0 C) = 11.0 C

m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg

11.0 = 1.86 * moles of methanol/4.50

moles of methanol = 26.613 moles

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Carbonic acid reacts with water to yield bicarbonate ions and hydronium ions: h2co3+h2o?hco3?+h3o+ identify the conjugate acid-b

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3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p

almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen 40300 g

Carbon 11160 g

Hydrogen 6200 g

Nitrogen 1860 g

Calcium 992 g

Phosphorus 744 g

Other elements 744 g

It is possible to check it adding all the components, the total is 62000 g.

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

Percentage ( %)

Oxygen 65.0

Carbon 18.0

Hydrogen 10.0

Nitrogen 3.0

Calcium 1.6

Phosphorus 1.2

Other elements 1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

Percentage ( %) Decimal number

Oxygen 65.0 65/100 = 0.65

Carbon 18.0 18/100 = 0.18

Hydrogen 10.0 10/100 = 0.10

Nitrogen 3.0 3/100 = 0.03

Calcium 1.6 1.6/100 = 0.016

Phosphorus 1.2 1.2/100 = 0.012

Other elements 1.2 1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

Percentage ( %) Decimal number Mass in grams (g)

Oxygen 65.0 65/100 = 0.65 0.65*62000=40300

Carbon 18.0 18/100 = 0.18 0.18*62000= 11160

Hydrogen 10.0 10/100 = 0.10 0.10*62000= 6200

Nitrogen 3.0 3/100 = 0.03 0.03*62000= 1860

Calcium 1.6 1.6/100 = 0.016 0.016*62000= 992

Phosphorus 1.2 1.2/100 = 0.012 0.012*62000=744

Other elements 1.2 1.2/100 = 0.012 0.012*62000=744

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3 years ago