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3 years ago

What is the formal charge of nitrogen in no2cl

Chemistry

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

The formal charge on nitrogen in $NO_{2}Cl$ is +1.

Explanation:

The structure of $NO_{2}Cl$ is as follows.

(In attachment)

$Formal\, charge = Valence\, electron - (nonbonding\, electrons + \frac{Bonding\,electrons}{2})$

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.

$Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1$

Therefore, The formal charge on nitrogen in $NO_{2}Cl$ is +1.

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What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

⠀

The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

To find molality, we have to divide no. of moles of solute by weight of solution

⠀

While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

⠀

Now, we will find molality

⠀

$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

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Henceforth, the change in boiling point is 0.00735°C.

7 0

2 years ago

Help needed urgently the word missing or partially seen is oxygen I forgot to put more pointa

butalik [34]

Explanation:

2.04 % hydrogen

32.65% sulphur

65.31% is oxygen

atomic ratio

hydrogen =2.04÷1=2.04

sulphur =32.65÷32=1.02

oxygen =65.31÷16=4.08

simplest ratio

hydrogen = 2.04÷1.02=2

sulphur =1.02÷1.02=1

oxygen =4.08÷1.02=4

empirical formula is H2SO4

4 0

3 years ago

DerKrebs [107]

Answer:

F = 30 N

Explanation:

Given data:

Mass of toy train = 1.5 kg

Acceleration of train = 20 m/s²

Amount of force acting on it = ?

Solution:

The net force on object is equal to the its mass multiply by its acceleration.

Formula:

F = ma

F = force

m = mass

a = acceleration

Now we will put the values in formula.

F = 1.5 kg × 20ms⁻²

F = 30 kg.ms⁻²

kg.ms⁻² = N

F = 30 N

7 0

3 years ago

Identify the half reaction as oxidation or reduction.

topjm [15]

OILRIG:
Oxidation is loss (of electrons)
Reduction is gain (of electrons)
so...
The first one is an oxidation half-equation as the Sn loses electrons;
The second one is a reduction half-equation as the Cl₂ gains electrons

8 0

3 years ago

A 100.5 ml intraveneous (iv) solution contains 5.10 g glucose (c6h12o6). what is the molarity of this solution?

Pavlova-9 [17]

Answer is: the molarity of this glucose solution is 0.278 M.
m(C₆H₁₂O₆) = 5.10 g.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆) .
n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆) = 0.028 mol.
V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆) = 0.278 mol/L.

8 0

3 years ago