27. How was the design of the Chernobyl plant different than most modern reactors?

What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?

Airida [17]

Answer:

$V_1=0.0714L$

Explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

$n_1=n_2$

And in terms of volume and concentration:

$C_1V_1=C_2V_2$

Thus, we can solve for the volume of the concentrated HCl as shown below:

$V_1=\frac{C_2V_2}{C_1}$

Therefore, we plug in the data to get:

$V_1=\frac{0.250M*0.500L}{1.75 M}\\\\V_1=0.0714L$

Best regards!

7 0

3 years ago

Calculate the volume in liters of a M potassium dichromate solution that contains of potassium dichromate . Round your answer to

maksim [4K]

The question is incomplete, here is the complete question:

Calculate the volume in liters of a 0.13 M potassium dichromate solution that contains 200. g of potassium dichromate . Round your answer to 2 significant digits.

Answer: The volume of solution is 5.2 L

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Molarity of solution = 0.13 M

Given mass of potassium dichromate = 200. g

Molar mass of potassium dichromate = 294.15 g/mol

Putting values in above equation, we get:

$0.13M=\frac{200}{294.15\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{294.15\times 0.13}=5.23L$

Hence, the volume of solution is 5.2 L

3 0

3 years ago

If you complete and balance the following oxidation-reduction reaction in basic solution NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(a

andreev551 [17]

Answer:

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

Step 2: Balancing electrons

NO2- → NH3

On the left side N has an oxidation number of +3 and on the right side -3.

NO2- +6e-→ NH3

Al(s) → Al(OH)4-

On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

NO2- +6e-→ NH3

2(Al(s) → Al(OH)4- +6e-)

Step 3: Balance with OH/H2O

NO2- +6e +5H2O → NH3 +7OH-

2Al +8OH- → 2Al(OH)4- + 6e-

Step 4: The netto reaction

NO2- + 5H2O + 2Al + 8OH- → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

6 0

4 years ago

Give the name of the element that is a member of the alkali metal family whose most stable ion contains 2 electrons.

notka56 [123]

Lithium

Li → Li⁺ + e⁻

Li 1s²2s¹ → Li⁺ 1s² + e⁻

3 0

4 years ago

20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equ

Romashka-Z-Leto [24]

Answer:

The amount of Cl2 gas left , after the reaction goes to completion is : 139.655 grams

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

$Moles=\frac{Mass}{Molar\ mass}$

$2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)$

According to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

$Moles=\frac{Mass}{Molar\ mass}$

$Moles=\frac{20.1}{26.98}$

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

$Moles=\frac{Mass}{Molar\ mass}$

$Moles=\frac{219}{70.98}$

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2 mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

Cl2 in grams = 1.964 x 70.90 = 139.655 grams

6 0

4 years ago