All categories

3 years ago

Sodium chloride (table salt,) whose formula is NaCl: Ionic Covalent

Chemistry

2 answers:

Vesna [10]3 years ago

7 0

Since sodium chloride contains both a metal AND a nonmetal, the combination of those would result in an ionic bond.

lys-0071 [83]3 years ago

6 0

Answer: Ionic

Explanation:

Ionic bonds are formed by transfer of electrons between metal and non metals. Covalent bonds are formed by sharing of electrons between non metals

In Sodium chloride

Electronic configuration of sodium:

$[Na]=1s^22s^22p^63s^1$

Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.

$[Na^+]=1s^22s^22p^63s^0$

Electronic configuration of chlorine:

$[Cl]=1s^22s^22p^63s^23p^5$

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

$[Cl^-]=1s^22s^22p^63s^23p^6$

In sodium chloride the one electron from sodium metal gets transferred to chlorine atom and thus is ionic.

You might be interested in

What is the identity of the element that has 19 protons and 22 neutrons?

Degger [83]

Answer:

A titanium ion

3 0

3 years ago

The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.2

Rama09 [41]

Answer:

$-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}$

Explanation:

The rate law of a chemical reaction is given by

$-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}$

This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found

Between experiments 1 and 2

$\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta$

Then the expression for the calculation of $\beta$

$\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}$

Resolving

$\beta=1$

Doing the same between experiments 3 and 4 the expression for $\alpha$ is

$\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}$

Resolving

$\alpha=1$

This means that the rate law for this reaction is

$-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}$

5 0

4 years ago

A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th

Ann [662]

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

Q = M x C x ΔT

Q = -100 x 4.18 x (-5.14)

Q = 2148.5 joules

Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

ΔH = Q / number of moles of NaHCO3

ΔH = 2.1485 / 0.173

ΔH = 12.42 KJ/mol

3 0

3 years ago

Whats 2×2 rhdbdjdjddjdjkdjdbdbfbdbxb

Elan Coil [88]

Answer:

The answer might be 4, been a while since I've done third grade math though.

8 0

3 years ago

Read 2 more answers

At this temperature, 0.300 mol H 2 0.300 mol H2 and 0.300 mol I 2 0.300 mol I2 were placed in a 1.00 L container to react. What

Alekssandra [29.7K]

Answer:

The concentration of HI present at equilibrium is 0.471 M.

Explanation:

5 0

3 years ago