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3 years ago

When sodium chloride reacts with iron it forms sodium and iron (ll) Chloride. Write down the balanced equation and classify it \

Chemistry

1 answer:

BartSMP [9]3 years ago

7 0

Answer:

NaCl + Fe2+ ---> FeCl2 + Na+

Single replacement

Explanation:

Na is replaced by Fe2+ to form FeCl2, and that is the only replacement. Thus, it is single replacement.

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Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti

Lina20 [59]

Answer: The volume of stock solution needed is 90 mL

Explanation:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=6.0M\\V_1=?mL\\M_2=1.2M\\V_2=450mL$

Putting values in above equation, we get:

$6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL$

Hence, the volume of stock solution needed is 90 mL

6 0

2 years ago

Write a double replacement precipitation reaction A precipitate forms when aqueous solutions of nickel(II) chloride and silver(I

harkovskaia [24]

Answer: The balanced chemical equation is written below.

Explanation:

Double displacement reaction is defined as the reaction in which exchange of ions takes place.

$AB+CD\rightarrow CB+AD$

Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

When nickel (II) chloride reacts with silver (I) nitrate, it leads to the formation of white precipitate of silver chloride and an aqueous solution of nickel (II) nitrate.

The balanced chemical equation for the above reaction follows:

$NiCl_2(aq.)+2AgNO_3(aq.)\rightarrow 2AgCl(s)+Ni(NO_3)_2(aq.)$

By Stoichiometry of the reaction:

1 mole of aqueous solution of nickel (II) chloride reacts with 2 moles of aqueous solution of silver (I) nitrate to produce 2 moles of solid silver chloride and 1 moles of aqueous solution of nickel (II) nitrate

Hence, the balanced chemical equation is written above.

3 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

7 0

3 years ago

How many grams is 10 moles of H2O?

Elan Coil [88]

Answer:

So, 10 mole of water will weigh (18x10) = 180g.

5 0

2 years ago

Write the formula of the coordination compound pentaamminecarbonatocobalt(III) iodide. Enclose the coordination complex in squar

7nadin3 [17]

Answer:

[Co(NH3)5CO3]I3

Explanation:

The naming of coordination compounds follows certain rules specified by IUPAC. Usually, the name of the complex makes it quite easy to deduce its structure.

"Pentaamine" means that there are five NH3 ligands as shown in the structure. The ligand carbonato is CO3^2-. It has no prefix attached to it in the IUPAC name of the complex hence there is only one carbonato ligand present(recall that the complex has a coordination number of six). I did not enclose it within parenthesis as required in the question.

Lastly the III that appeared after the metal name "cobalt" shows its oxidation state. The iodide counter ions must then be 3 in number in order to satisfy this primary valency of the metal hence the inclusion of I3 in the structure of the complex.

6 0

3 years ago