Gizmo Covalent Bonds

Which of these has the greatest compressibility?

oksian1 [2.3K]

Gas is the most compressible because the atoms have more space in between. solids on the other hand, are the least compressible because their atoms are really closely packed.

7 0

3 years ago

Based on nuclear stability, what is the symbol for the most likely product nuclide when iodine-123 undergoes decay?

MissTica

Answer is: Te (tellurium).
During electron capture, iodine-123 form the nearly-stable nuclide tellurium-123.
In beta plus decay (atomic number Z is decreased by one), a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.
Iodine-123 is a radioactive isotope of iodine used in nuclear medicine imaging.

6 0

3 years ago

A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ

navik [9.2K]

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol$

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

$0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol$

For the given chemical equation:

$2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)$

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

$0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g$

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

5 0

3 years ago

Read 2 more answers

What are 4 other gases in the atmosphere besides oxegen and nitergin?

Levart [38]

Helium, neon,nitrogen and argon

5 0

3 years ago

The equilibrium constant is given for one of the reactions below.

Cerrena [4.2K]

Answer:

12.7551

Explanation:

The given chemical equation follows:

$2HD(g)\rightarrow H_2(g)+D_2(g)$

The equilibrium constant for the above equation is 0.28.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$2H_2(g)+2D_2(g)\rightarrow 4HD(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(\frac{1}{0.28})^2$

Hence, the value of equilibrium constant for reverse reaction is 12.7551.

3 0

4 years ago