Please help!!

A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of

blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol$

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

0.01 mol of $H^+$ ion will react with = $\frac{1}{1}\times 0.01mole$ of $OH^-$ ion

Thus (0.012-0.01)= 0.002 moles of $OH^-$ are left in 20 ml or 0.02 L of solution.

$[OH^-]=\frac{0.002}{0.02L}=0.1M$

$pOH=-log[OH^-]$

$pOH=-log[0.1]=1$

$pH+pOH=14$

$pH=14-1=13$

Thus the pH of resulting solution will be 13

7 0

2 years ago

What is most likely to happen when water loses heat?

ANTONII [103]

A

Hope this helped thanks u

3 0

3 years ago

The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations.

lions [1.4K]

Answer:

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

Explanation:

The osmotic pressure (π) is the pressure needed to impede the osmose, it means that it's the necessary pressure to prevent the solvent to go through a membrane.

It can be calculated by:

π = M*R*T*i

Where M is the molarity of the solution (mol/L), R is the ideal gas constant, T is the temperature, and i the van't Hoff factor.

This factor is a way to correct the number of particles that are dissolved in a solute, and it can be calculated by:

i = 1 + α*(q - 1)

Where α is the degree of dissociation of a substance, and q is the number of moles of each ion released in a solution. Thus, covalent compounds that didn't ionize, such as sugars, have only one particle, and q = 1, and so i =1.

Because all the substances have the same molarity (1 M) and are at the same temperature, let's analyze the value of i, which is directly proportional to π.

C₁₂H₂₂O₁₁ is a sugar that didn't ionize, so π = 1;

Both KCl and MgCl₂ are soluble salts and will dissociate completely (α = 1), but MgCl₂ will have 3 particles (Mg²⁺ + 2Cl⁻), and KCl only one particle (K⁺ and Cl⁻), so qMgCl₂ > 1KCl, and so πMgCl₂ > πKCl, which will be higher than 1.

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

5 0

3 years ago

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the unw

Tomtit [17]

Answer:

0.1082M of Barium Hydroxide

Explanation:

KHP reacts with Ba(OH)2 as follows:

2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻

<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>

<em />

To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:

<em>Moles KHP -Molar mass: 204.22g/mol-</em>

1.37g * (1mol / 204.22g) = 0.006708 moles KHP

<em>Moles Ba(OH)2:</em>

0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =

0.003354 moles Ba(OH)2

<em>Molarity:</em>

0.003354 moles Ba(OH)2 / 0.0310L =

<h3>0.1082M of Barium Hydroxide</h3>

4 0

3 years ago

Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)

vampirchik [111]

Answer: Oxygen is the limiting reagent.

Explanation: $CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)$

As can be seen from the given balanced equation:

3 moles of $O_2$ reacts with 1 mole of $CS_2$

1.52 moles of $O_2$ reacts with= $\frac{1}{3}\times 1.52=0.51moles$ of $CS_2$

Thus $O_2$ is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of $CS_2$ will remain as such and thus $CS_2$ is an excess reagent.

6 0

3 years ago