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Inessa [10]
3 years ago
6

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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Given the following equation: Cu + 2 AgNO3 ---&gt; Cu(NO3)2 + 2
Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

  • Molar mass of AgNO3 = 169.87 g/mol
  • Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0
3 years ago
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices
BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0
3 years ago
Does anybody know the triple bond?
kkurt [141]

Answer:

a chemical bond in which three pairs of electrons are shared between two atoms

Explanation:

7 0
3 years ago
Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol
stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point):  K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

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We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

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b. CaCl₂ →  Ca²⁺  +  2Cl⁻      i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³     i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0
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Lady_Fox [76]

Answer:

B

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3 0
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