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A sample of solid calcium hdroxide, Ca(OH)2 is allowed to stand in water until a saturated solution is formed. A titration of 75.00mL of this solution with 5.00 x 10-2 M HCl 36.6 mL of the acid to reach the end pointCa(OH)2 + 2HCl ? CaCl + 2H2O
What is the molarity?

Answers

Answer 1

Answer:

Answer: The concentration of $Ca(OH)_2$ is 0.0122 M.

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\nM_1=5.00* 10^(-2)M=0.05M\nV_1=36.6mL\nn_2=2\nM_2=?M\nV_2=75mL$

Putting values in above equation, we get:

$1* 0.05* 36.6=2* M_2* 75\n\nM_2=0.0122M$

Hence, the concentration of $Ca(OH)_2$ is 0.0122 M.

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True or False: Observations are just things that you see

Answers

Answer:

False

Explanation:

"Observations is the action of observing something or someone carefully or in order to gain information" - google dictionary. This definition does not completely match the definition given, so the answer is false.

I hope this helps!

Answer: true

Explanation: the action or process of observing something or someone carefully or in order to gain information that is the real definition

Determine the (H+), pH, and pOH of a solution with an [OH-] of 9.5 x 10-10 M at 25 °C. M pH =

Answers

Answer : The concentration of $H^+$ ion, pH and pOH of solution is, $1.05* 10^(-5)M$ , 4.98 and 9.02 respectively.

Explanation : Given,

Concentration of $OH^-$ ion = $9.5* 10^(-10)M$

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pOH=-\log [OH^-]$

$pOH=-\log (9.5* 10^(-10))$

$pOH=9.02$

The pOH of the solution is, 9.02

Now we have to calculate the pH.

$pH+pOH=14\n\npH=14-pOH\n\npH=14-9.02=4.98$

The pH of the solution is, 4.98

Now we have to calculate the $H^+$ concentration.

$pH=-\log [H^+]$

$4.98=-\log [H^+]$

$[H^+]=1.05* 10^(-5)M$

The $H^+$ concentration is, $1.05* 10^(-5)M$

Answer:

pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978

Explanation:

Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C

As, pOH = - log [OH⁻]

⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022

The self-ionisation constant of water is given by

Kw = [H⁺] [OH⁻] and pKw = pH + pOH

Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.

Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M

also,

pH + pOH = pKw = 14

⇒ pH = 14 - pOH = 14 - 9.022 = 4.978

The quantity of mass of an object contained within its volume is a measure of

Answers

the answer is density

the quantity of mass of an object contained within its volume is a measure of density

The most common carbonate rock is A) dolomite. B) halite. 35) C) limestone. D) calcite.

Answers

Answer:

The correct option is : C) limestone.

Explanation:

Carbonate rocks are a type of sedimentary rocks. The carbonate rocks are composed of carbonate minerals. The carbonate minerals are the minerals containing carbonate ion (CO₃²⁻).

The most common type of carbonate rock is limestone. Limestone is composed of the minerals calcite and aragonite, which have a different crystal form of calcium carbonate.

Therefore, Limestone is the most common type of carbonate rock.

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1,123 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -340 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -211 kJ