What natural resources are used to make plastic bags

If the total number of H30+ ions are greater than the OH- the resulting solution will be?

Solnce55 [7]

Answer:

acidic

Explanation:

The number of H+ and OH- ions are how pH is calculated

since pH is -log [H+] more H+ ions create a more acidic solution

7 0

3 years ago

2HCl+Ca(OH)2⟶CaCl2+2H2O

RUDIKE [14]

Isn’t it just 3 mols?

5 0

3 years ago

Calculate the molarity of an hcl solution if a 10.00 ml sample requires 25.24 ml of a 1.600 m naoh solution to be neutralized.

AlladinOne [14]

Answer:

The molarity of the HCl solution should be 4.04 M

Explanation:

Step 1: Data given

volume of HCl solution = 10.00 mL = 0.01 L

volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L

Step 2: The balanced equation

HCl + NaOH → NaCL + H2O

Step 3: Calculate molarity of HCl

n1*C1*V1 = n2*C2*V2

Since the mole ratio for HCl and NaOH is 1:1 we can just write:

C1*V1 =C2*V2

⇒ with C1 : the molarity of HCl = TO BE DETERMINED

⇒ with V1 = the volume og HCl = 10 mL = 0.01 L

⇒ with C2 = The molarity of NaOH = 1.6 M

⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L

C1 * 0.01 = 1.6 * 0.02524

C1 = (1.6*0.02524)/0.01

C1 = 4.04M

The molarity of the HCl solution should be 4.04 M

6 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago

Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.

Deffense [45]

Answer 1) When a strong acid like $HNO_{3}$ reacts with $Ag_{2} CO_{3}$ usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when $HNO_{3}$ reacts with $Ag_{2} CO_{3}$ it generates carbon dioxide and water as a by product, if we are adding $HNO_{3}$ it will remove some of the $CO_{3}$ molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace $HNO_{3}$ with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing $Ag^{+}$ and $Cl^{-}$ ions from the solution.

7 0

3 years ago