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3 years ago

How does changing the ratio of nitrogen atoms and oxygen atoms change the compounds

Chemistry

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

Answer:

Explanation:

Compounds consist of atoms of different elements. In compound the atoms are present in fixed proportion. By changing the proportion compound will changed. For example,

Nitrogen and oxygen reacted to form the compound. By changing the proportion both will form different compound.

NO₂

Nitrogen dioxide consist of one atom of nitrogen and two atoms of oxygen. Their ration is 1 :2

In nitric oxide one atom of oxygen and one atom of nitrogen are present.

Their ratio is 1 : 1.

N₂O

Nitrous oxide consist of two nitrogen atoms and one oxygen. Their ratio is

2 : 1

N₂O₄

Dinitrogen tetroxide consist of two nitrogen atoms and four oxygen atoms. Their ratio will written as 2 : 4

N₂O₅

Dinitrogen pentoxide consist of two nitrogen atoms and five oxygen atoms. Their ratio will written as 2 : 5

It can be seen that by changing the ratio of atoms compounds are changed.

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An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.

slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

$m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg$

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

$m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg$

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

A chlorine atom has a diameter of 175 picometers (pm). Please write the diameter, in metered, in scientific notation.

Flauer [41]

Picometre is equal to 1·10⁻¹²m
175·10⁻¹² m = 1,75·10⁻¹⁰ m

7 0

3 years ago

I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:

dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios. This makes up for using the units that wouldn't appropriate elsewhere.

2. Use the equation P₁V₁ = P₂V₂. Solve for V₂.

(3.05 L)(870 kPa) = (969 kPa)(V₂)

V₂ = 2.74 L.

3. Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.

(3.32 L)/(360 K) = (4.50 L)/(T₂)

T₂ = 488 K.

4. Do the same as above, but for V₂.

(5.10 L)/(-56°C) = V₂/(-82°C)

V₂ = 7.47 L

5. Use the equation V₁/n₁ = V₂/n₂. Solve for V₂.

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)

V₂ = 38.6 L

6. Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)

n₂ = 2.85 mol

7. Use the equation P₁/T₁ = P₂/T₂. Solve for T₂.

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)

T₂ = 319 K

8. Do the same as above, but for P₂.

In this specific case, however you will need to convert units. Since both temperatures don't have the same sign, the ratio won't come out right. Convert to Kelvin. Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K 25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0

3 years ago