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2 years ago

Dot structure for HNF2

Chemistry

1 answer:

dimulka [17.4K]2 years ago

8 0

The total number of valence electrons is 20.(1 from hydrogen, 5 from nitrogen, and 7 from each fluorine so 14 for both) connect all the atoms with a single bond( each bond counts as 2) so far there is 6 electrons so you need to fill in the rest of the 14 and you get the result in the picture.

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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

2 years ago

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

Citrus2011 [14]

Depression in freezing point (Δ $T_{f}$ ) = $K_{f}$ ×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$ ,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$ )

Thus, (Δ $T_{f}$ ) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ $T_{f}$ ) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ $T_{f}$ ) = - $0.03162^{0}C$

8 0

2 years ago

Read 2 more answers

The crystalline hydrate cd(no3)2 ⋅ 4h2o(s) loses water when placed in a large, closed, dry vessel at room temperature: cd(no3)2⋅

Sever21 [200]

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

8 0

2 years ago

Read 2 more answers

Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

$m_{tot}=m_{fuselage}+m_{passangers}$

For aluminium:

$m_{tot}=m_{fus-Al}+m_{pas-Al}$

$m_{fus-Al}=\delta _{Al}*V_{fuselage}$

and

$V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]$

where:

L is lenght
D is diameter
e is thickness

$m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}$

For steel (same procedure):

$m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel$

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0

2 years ago

What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its

pishuonlain [190]

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT = T2 - T1

ΔT = 20°C - 25°C

ΔT = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

50208 j /-12.5 g.°C = c

c = 4016.64 j/g.°C

7 0

2 years ago

Dot structure for HNF2​

Dot structure for HNF2