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2 years ago

In the correct lewis structure for the nitrate ion, no3 –, nitrogen has how many covalent bonds?

Chemistry

1 answer:

Colt1911 [192]2 years ago

8 0

Answer : The correct Lewis-dot structure of nitrate ion is shown below and the nitrogen has 4 covalent bonds.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule. The number of valance electrons are shown by 'dot'.

The given molecule is, nitrate ion

As we know that nitrogen has '5' valence electrons and oxygen has '6' valence electron.

Therefore, the total number of valence electrons in $NO_3^-$ = 5 + 3(6) + 1 = 24 electrons

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

The correct Lewis-dot structure of nitrate ion is shown below.

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On top of one of the peaks in rocky mountain national park the pressure of the atmosphere is 550 torr determine the boiling poin

Dahasolnce [82]

Answer:

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

Explanation:

Step 1: Data given

Pressure = 550 torr

The heat of vaporization of water is 40.7 kJ/mol.

Step 2: Calculate boiling point

⇒ We'll use the Clausius-Clapeyron equation

ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)

ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)

⇒ with P1 = 760 torr = 1 atm

⇒ with P2 = 550 torr

⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin

⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED

ln(550/760) = 4895.4*(1/373.15 - 1/T2)

-0.3234 = 13.119 - 4895.4/T2

-13.4424= -4895.4/T2

T2 = 364.2 Kelvin = 91 °C

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

4 0

1 year ago

The hydrogen and oxygen atoms of a water molecule are held together by ________ bonds. The hydrogen and oxygen atoms of a water

Aleksandr [31]

Answer: HYDROGEN BONDS

Explanation:

Water molecules attract each other happily thanks to their polarity. A hydrogen atom plus end associates an oxygen atom minus end.

These attractions are an example of hydrogen bonds, weak interactions forming between a partially positive charged hydrogen and a more electronegative atom like oxygen. The hydrogen atoms involved in bonding with hydrogen need to be bound to electronegative atoms such as Oxygen and fluorine

3 0

1 year ago

Read 2 more answers

Determine the number of moles and mass requested for each reaction in Exercise 4.42.

suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

Step 1: Write the balanced equation
Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$ [/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$

5 0

1 year ago

A 2400.-gram sample of an aqueous solution contains 0.012 gram of nh3. What is the concentration of nh3 in the solution

Kryger [21]

Answer : The concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

Explanation :

First we have to calculate the volume of aqueous solution that is water.

Density of water = 1.00 g/mL

Mass of water = 2400 g

$\text{Volume of water}=\frac{Mass}{Density}=\frac{2400g}{1.00g/mL}=2400mL$

Now we have to calculate the concentration of ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }NH_3\times 1000}{\text{Molar mass of }NH_3\times \text{Volume of solution (in mL)}}$

Molar mass of $NH_3$ = 17 g/mole

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.012g\times 1000}{17g/mole\times 2400mL}=3.12mole/L=2.94\times 10^{-4}M$

Therefore, the concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

7 0

2 years ago

By which process is a precipitate most easily separated from the liquid in which it is suspended

Gelneren [198K]

Filtration, if its a precipitate that means its insoluble.

7 0

1 year ago

Read 2 more answers

In the correct lewis structure for the nitrate ion, no3 –­, nitrogen has how many covalent bonds?

In the correct lewis structure for the nitrate ion, no3 –, nitrogen has how many covalent bonds?