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2 years ago

explain how energy changes from one form to another in a exothermic reaction. in an endothermic reaction.

Chemistry

1 answer:

Alexxx [7]2 years ago

6 0

Answer:

Exothermic reaction: In exothermic reaction, energy is transferred to the surroundings, and the surrounding temperature increases, this is known as exothermic reaction. In other words energy exits in exothermic reaction. Some example of exothermic reactions are:

1) Neutralisation reaction.

2) Combustion reaction.

3) Some oxidation reaction.

Endothermic reaction: In endothermic reaction, energy is taken in from the surrounding, and the surrounding temperature decreases, this is known as endothermic reaction. In other words energy enters in endothermic reaction. Some example of exothermic reactions are:

1) Thermal decomposition.

2) Reaction between citric acid and sodium hydrogen carbonate.

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A. Consider four different samples: aqueous LiI, molten LiI, aqueous AgI, and molten AgI. Current run through each sample produc

Rudiy27

Answer:

Explanation:

At the cathode

In case of molten AgI

Silver will be collected

In case of molten LiI

lithium will be collected

in case of aqueous LiI,

hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺

in case of aqueous AgI,

Silver will be obtained at cathode because reduction potential of silver is more than H⁺

At the Anode

In case of molten NaBr

Bromine will be collected

In case of molten NaF

Fluorine will be collected

in case of aqueous NaBr ,

Bromine will be collected as reduction potential of Br⁻ is less than O⁻²

in case of aqueous NaF ,

oxygen will be obtained because reduction potential of F⁻ is more than O⁻² .

5 0

2 years ago

The weight percent of concentrated HClO4(aq) is 70.5% and its density is 1.67 g/mL. What is the molarity of concentrated HClO4

ollegr [7]

Answer:

[HClO₄] = 11.7M

Explanation:

First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.

Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles

Now we can apply the density to calculate the volume.

Density always refers to solution → Solution density = Solution mass / Solution volume

1.67 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.67 g/mL → 59.8 mL

To determine molarity (mol/L) we must convert the mL to L

59.8 mL . 1L/1000mL = 0.0598 L

Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M

8 0

2 years ago

kondor19780726 [428]

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

$\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0

2 years ago

Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give

denpristay [2]

Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T

Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J

T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K

3 0

2 years ago

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p

Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0

2 years ago

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