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kow [346]
3 years ago
10

What is the pH of a Koh solution that has [H+]=1.87×10^-13M

Chemistry
2 answers:
xenn [34]3 years ago
7 0
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72

I hope this helps.  Let me know if anything is unclear.
vodomira [7]3 years ago
6 0

Answer:

12.7

Explanation:

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In the partition coefficient experiment 4A this week, after thorough mixing of the reagents, phase separation will occur. The to
Andreas93 [3]

Answer:

The top layer is the Aqueous layer, and the benzoic acid is contained in the non-aqueous layer/oil phase.

Explanation:

A separating funnel is a very important piece of laboratory glassware that is used to separate the components of liquid-liquid mixtures  which are immiscible. This technique is used in the extraction of the components of mixtures.

The liquids separate into two phases. The separation is based on the differences in the liquids' densities, where the denser liquid settles below and the lower density liquid stays afloat. Liquids used for this kind of separation are usually different liquids, one is the aqueous layer and the other, a non-aqueous layer.

Partition coefficient or distribution coefficient is defined as the ratio of the concentrations of a compound in two immiscible solvents at equilibrium.

Organic solvents (except halogenated organic compounds) with densities greater than that of water i.e 1g/mL ( usually called the oil phase) settle at the bottom of the aqueous phase.

Benzoic acid. will settle at the bottom layer ( i.e the lower phase).

7 0
3 years ago
Which statement explains whether NaCl or BeO will have a stronger bond?
Alex777 [14]

Answer:

C

Explanation:

8 0
3 years ago
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0
3 years ago
Determine the [h3o+] of a 0.210 m solution of formic acid.
Nataly [62]
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka 
   3.77 = -㏒ Ka
∴Ka = 1.7x10^-4 

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2      by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
       = 2.23 

3 0
3 years ago
What volume does 0.482 mol of gas occupy at a pressure of 719 mmHg at 56 ∘C?
kozerog [31]

Answer:

The volume is 13, 69 L

Explanation:

We use the formula PV=nRT. We convert the temperature in Celsius into Kelvin and the pressure in mmHg into atm.

0°C= 273K---> 56°C= 56 + 273= 329K

760 mmHg----1 atm

719 mmHg----x= (719 mmHgx 1 atm)/760 mmHg= 0,95 atm

PV=nRT ---> V= (nRT)/P

V=( 0,482 molx 0,082 l atm/K mol x 329K)/0,95 atm

<em>V=13,68778526 L</em>

7 0
3 years ago
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