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swat32
3 years ago
6

What type of reaction is: sulfuric acid + potassium hydroxide -> potassium sulfate + water?

Chemistry
1 answer:
aleksley [76]3 years ago
4 0

Answer:

Double replacement reaction

Explanation:

Now, let us first write the reaction equation properly:

     H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O

The above reaction is a neutralization reaction between an acid and a base whose product gives salt and water only at most instances.

From here, we can observe that the species displaces on another in their ionic state. Hydrogen replaces potassium and water is produced. Potassium combines chemically with sulfate ions to give the salt of potassium.

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An ice cube is placed in 80 mL of 70oC water. When the ice melts the temperature of the water is 60oC. a. How much energy was ne
tigry1 [53]

Answer:

800 degrees C g, I found this answer by using the formula to find how much energy was need to melt the ice. First you figure out the temperature change of ur ice (starting temp and ur final temp), and then you times that with the mass of the water.

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2 years ago
Why is mercury the only metal to have been used in thermometers
MArishka [77]
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3 years ago
Pls help me. Show proof on why it’s that answer.
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2 years ago
How is a molecule of hydrogen formed from two hydrogen atoms?
Shtirlitz [24]
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8 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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