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3 years ago

Which of the following is a correct observation when water is added to quick lime?

Chemistry

2 answers:

masya89 [10]3 years ago

7 0

Answer:

Explanation:

thw observation of water that is added

il63 [147K]3 years ago

3 0

D)no change
Cause water is a poor conductor of heat and electricity

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A) How many electron pairs are on the central atom of an H2S molecule?

zlopas [31]

Answer:

Each Hydrogen atom has one only one electron which is also its valence electron and Sulfur has six valence electrons. Thus, there are a total of eight valence electrons in H2S. In this compound, both the hydrogen atoms require one electron to make the covalent bond with Sulfur

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3 years ago

NH4+H2O →NH3+H3O Which one is acting weaker base?

Nataliya [291]

Answer:

NH3

Explanation:

In solution, they are in equilibrium. NH4+ acts as a bronsted Lowry acid and donates an H to become NH3, and NH3 acts as a bronsted lowry base and accepts an H. In this pair, NH3 is a weak base, which gets its basic character due to the presence of lone pair of nitrogen and its ability to donate it.

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2 years ago

Energy from the Sun reaches us via: * Radiation Conduction Convection Reflection

kramer

Answer:

the andswer is radiation

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3 years ago

The______ the forces among the particles in a sample of matter, the more rigid the matter will be.

Alexus [3.1K]

Answer:

Stronger!

Explanation:

The stronger the forces among the particles in a sample of matter, the more rigid the matter will be.

3 0

4 years ago

A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate (a) the mass

IgorLugansk [536]

Answer:

a) 43%

b) 0.122

c) 7.70 molal

d) 5.83 M

Explanation:

Step 1: Data given

Mass of H2SO4 = 571.6 grams

Density of H2SO4 = 1.329 g/cm³ this means 1.329 grams per 1 mL or 1329 grams per 1L

a) Calculate mass percentage

Mass % = (571.6 grams / 1329 grams)* 100% = 43%

b) Calculate mole fraction

Number of moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

Moles H2SO4 = 571.6 grams / 98.08 g/mol

Moles H2SO4 = 5.83 moles

Moles H2O = (1329 -571.6)/18.02 = 42.03 moles

Total moles = 5.83 + 42.03 = 47.86 moles

Mole fraction H2SO4 = Moles of solute (H2SO4)/ Total moles

Mole fraction H2SO4 = 5.83 moles / 47.86 moles

Mol fraction h2SO4 = 0.122

c) Calculate the molality

Mass of solvent = 1329 grams - 571.6 grams = 757.4 grams = 0.7574 kg

Molality of H2SO4 = number of moles H2SO4 / mass of solvent

Molality H2SO4 = 5.83 moles / 0.7574 kg

Molality H2SO4 = 7.70 molal

d) Calculate Molarity

Molarity H2SO4 = Number of moles H2SO4 / volume

Molarity H2SO4 = 5.83 moles / 1L = 5.83 M

6 0

3 years ago