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2 years ago

Explain what is water of crystallization

Chemistry

1 answer:

ArbitrLikvidat [17]2 years ago

3 0

Explanation:

In chemistry, water(s) of crystallization or water(s) of hydration are water molecules that are present inside [crystal]s. Water is often incorporated in the formation of crystals from aqueous solutions. ... Water of crystallization can generally be removed by heating a sample but the crystalline properties are often lost

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What is answer and why

chubhunter [2.5K]

D because O3 means oxygen times 3 times it self so you add another oxygen partial it becomes O4

6 0

2 years ago

Determine the molarity of a solution prepared by dissolving 141.6 g of citric acid , c3h5o(cooh)3 , in water and then diluting t

gogolik [260]

The molarity of a solution prepared by dissolving 141.6g of citric acid in water is calculated as below

find the the number of moles

moles= mass/molar mass

= 141.6 g/ 192 g/mol = 0.738 moles

molarity= moles/molar mass

= 0.738/3500 x1000 = 0.21 M

3 0

3 years ago

Among the following, which element has the lowest ionization energy? Question 2 options: Cs Na Cl I

marissa [1.9K]

The element with the lowest ionization energy is CESIUM, CS.
Ionization energy is the energy required to remove the most loosely bound electron in an atom of an element. The higher the number of shells in an atom of an element, the lower the ionization energy that will be required to remove the valence electron.

7 0

2 years ago

Read 2 more answers

If you have a density of 100 kg/L and a mass of 1000 g what is the volume (please explain)

olchik [2.2K]

Answer: Volume = 0.01L

Explanation:

The density of a substance is given by; Density = Mass / Volume.

In this question, Mass = 1000g, Volume = ? and Density = 100 Kg/L

For the units to be uniform, we convert 1000g to Kg = 1Kg

Therefore, Volume = Mass / Density = 1Kg / 100Kg/L = 0.01L

6 0

3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

Explain what is water of crystallization​

Explain what is water of crystallization