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joja [24]
3 years ago
7

The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprot

ic acid with 3 pKa values: 2.14, 6.86, and 12.4.
Chemistry
1 answer:
Bezzdna [24]3 years ago
7 0

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

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mars1129 [50]
I believe that it most likely would be C,
6 0
3 years ago
28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?
lora16 [44]
The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

28.7 L C3H8 ( 1 mol / 22.41 L ) ( 4 mol H2O / 1 mol C3H8 ) ( 18.02 g / mol ) = 92.31 g H2O produced

Hope this answers the question.
7 0
3 years ago
The pOH of a solution is 9.21. Calculate the hydrogen ion concentration of the solution. Be sure to report your answer to the co
11Alexandr11 [23.1K]

Answer:

[OH-] = 6.17 *10^-10

Explanation:

Step 1: Data given

pOH = 9.21

Step 2: Calculate [OH-]

pOH = -log [OH-] = 9.21

[OH-] = 10^-9.21

[OH-] = 6.17 *10^-10

Step 3: Check if it's correct

pOH + pH = 14

[H+]*[OH-] = 10^-14

pH = 14 - 9.21 = 4.79

[H+] = 10^-4.79

[H+] = 1.62 *10^-5

6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14

3 0
3 years ago
Is butter solution or colloid or suspension
netineya [11]
Hi , butter is a class of colloids called emulsions , so your answer is colloid.
6 0
4 years ago
A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperatu
lapo4ka [179]

Answer:

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

6 0
3 years ago
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