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3 years ago

What is the conjugate base of the following acids:1. HCIO,2. PH4^+

Chemistry

1 answer:

Andrew [12]3 years ago

7 0

Answer:

ECUACIÓN:HClO 2 + H 2O → ClO− 2 + H 3O

ACIDO: HClO2

BASECONJUGADA:ClO-2

Explanation:

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You just measured a metal cylinder and obtained the following information: mass - 3.543 g diameter -0.53 cm height = 4.40 cm. 26

Fittoniya [83]

Answer:

V cylinder = $\pi r^2 h = \pi(0.53 cm/2)^24.4cm= 0.97 cm^3$

note that $r=diameter/2$

density = $m/V= 3.543/0.97= 3.65 g/cm^3$

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3 years ago

Can someone please help me with my science research I kind of need help in a few things that I listed, I’m currently working on

dalvyx [7]

Answer:

For your first question, Curium does not occur naturally on Earth, meaning that it is not produced naturally on Earth. However, it can be formed in nuclear reactors.

For your second question, Curium has been used to provide power to electrical equipment used on space missions, but doesn't seem to be that important overall.

Explanation:

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3 years ago

What kind of bond is ca-br

Genrish500 [490]

Answer: Ionic or electrovalent bond

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3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

4 years ago

A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The t

Leya [2.2K]

Answer:

12.99

Explanation:

<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

Mass of KOH: 716. mg (0.716 g)

Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

8 0

3 years ago

What is the conjugate base of the following acids:1. HCIO,2. PH4^+​

What is the conjugate base of the following acids:1. HCIO,2. PH4^+