Ask question

Ask question

All categories

3 years ago

5

If chlorine gas is bubbled through an aqueous solution of sodium iodide, the result is elemental iodine and aqueous sodium chlor

ide solution. what kind of reaction took place?

2 answers:

zvonat [6]3 years ago

3 0

The balanced reaction that describes the reaction of chlorine gas and sodium iodide to produce elemental iodine and sodium chloride in aqueous solution is expressed Cl2+2NaI= I2 + 2NaCl. This kind of reaction is called single replacement reaction where the anion, in this case, is only replaced.

Dmitrij [34]3 years ago

3 0

The correct answer is

Single replacement

:)

You might be interested in

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

$Molecular\ weight = \frac{Mass\ in\ g}{Moles}$

Molecular weight = $\frac{1.450}{0.00389} = 372.13\ g/mol$

4 0

3 years ago

Which list accurately identifies the hierarchy of space from largest to smallest?

katovenus [111]

D)Universe, galaxy, solar system, moon

4 0

3 years ago

mario62 [17]

Answer:

The correct answer is a the central core and is composed of protons and neutrons.

Explanation:

Nucleus is present in the center of an atom.Nucleus contain positively charged particle proton along with neutron which is neutral.

Due to the the presence of protons nucleus contain positive(+ve) charge.

4 0

3 years ago

A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde

hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is $\boxed{{\text{0}}{\text{.85}}}$ .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}$

......(1)

Rearrange equation (1) to calculate the moles of solute.

${\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)$ ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}$

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}$

The reaction between HCl and NaOH occurs as follows:

${\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}$

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

$\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}$

The volume after the addition of NaOH can be calculated as follows:

$\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}$

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

$\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}$

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

${\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]$ ......(3)

Here,

$\left[{{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of also becomes 0.141 M.

Substitute 0.141 M for $\left[{{{\text{H}}^ + }}\right]$ in equation (3).

$\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}$

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

4 0

3 years ago

Read 2 more answers

Clathrate hydrate research is done by Professor Kend Janda at UC Irvine. Look up clathrate hydrate with a general internet searc

lidiya [134]

Answer:

water, host

Explanation:

Clathrate hydrates are the crystalline water-based solids which physically resembles with the ice, in which the small non-polar molecules or smaller polar molecules with the large hydrophobic moieties are usually trapped inside the "cages" of the hydrogen bonded as well as frozen water molecules.

<u>Clathrate hydrates are the clathrate compounds in which water creates the cage and the water behaves as a host molecule and guest molecule is typically a gas or liquid.</u>

3 0

3 years ago