2. HBIO3 name this acid.

A student analyzed an unknown sample that contained a single anion. The sample gave a yellow precipitated upon addition of a sol

kenny6666 [7]

<h3>Answer:</h3>

Anion present- Iodide ion (I⁻)

Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)

<h3>Explanation:</h3>

In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.

Additionally we need to know the color of the precipitates.

Some of insoluble salts of silver and their color include;

Silver chloride (AgCl) - white color
Silver bromide (AgBr)- Pale cream color
Silver Iodide (AgI) - Yellow color
Silver hydroxide (Ag(OH)- Brown color

With that information we can identify the precipitate of silver formed and identify the anion present in the sample.

The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
Therefore, the anion that was present in the sample was iodide ion (I⁻).
Thus, the corresponding net ionic equation will be;

Ag⁺(aq) + I⁻(aq) → AgI(s)

4 0

3 years ago

Need answer asap 4 both

marishachu [46]

The persistence of vision is regarded as an optical illusion but it is actually part of human physiology.

<h3>What is persistence of vision?</h3>

The term persistence of vision is the idea that an object is able to remain is sight for a long time even when the object has been removed. This is often used by performers to create a remarkable impression on the audience. The persistence of vision is regarded as an optical illusion but it is actually part of human physiology.

The way that a performance can be designed to make the audience to know that this is the most important scene is called follow through.

Learn more about optical illusion:brainly.com/question/28179807

#SPJ1

7 0

1 year ago

Calculate the concentration of acetate ion in a buffer solution made from 2.00 mL of 0.50 M acetic acid and 8.00 mL of 0.50 sodi

Lelu [443]

Answer:

1 M

Explanation:

Equation of reaction is;

CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH

1 moles of each of the reactants react to give 2 moles of the acetate ion.

From the question, we have that 2.00 mL that is (2÷1000)L of 0.50 M acetic acid reacted with 8.00 mL that is (8/1000)L of 0.50 sodium acetate.

Then from equation, n = CV -------------------------------------------(1).

Where n= number of moles, V= volume, C= concentration.

Number of moles,n of acetic acid = 0.50M× 2/1000L.

n(acetic acid)= 0.001 moles.

Number of moles,n of sodium acetate= 0.50M ×(8/1000)L.

n(sodium acetate)= 0.004 moles.

0.001 moles of acetic acid react with 0.004 moles of Sodium acetate

Therefore, acetic acid is the limiting reagent.

One mole of acetic acid produces 2 moles of acetate ion.

0.001 mole of acetic acid produces= 0.002 moles of acetate ion.

Using the equation (1) that is, n= CV.

0.002= C× 2/1000

C= 0.002/0.002

C= 1 M

8 0

3 years ago

Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.)

rosijanka [135]

When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-

when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M

3 0

4 years ago

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit

pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$ .

How many moles of this process?

Relative atomic mass from a modern periodic table:

K: 39.098;
N: 14.007;
O: 15.999.

Molar mass of $\text{KNO}_3$ :

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$ .

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$ . By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$ .

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0

3 years ago

2. HBIO3 name this acid.​

2. HBIO3 name this acid.