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Bumek [7]
3 years ago
11

2. Which sentence explains how air pollution is harmful? *

Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:Air Pollution is when gas and fuel is released into the air and damages the environment. The Answer out of your answers would be C.

Explanation:

You might be interested in
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

brainly.com/question/5976752

brainly.com/question/13707449

brainly.com/question/13707449

brainly.com/question/6263007

brainly.com/question/14641878

brainly.com/question/2912965

#SPJ1

7 0
2 years ago
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
3 years ago
Help ASAP If you know what to do comment if u don’t I’ll report you ! Points added
mylen [45]

Proton:

Positive

Found in Nucleus

Mass of 1 AMU

Neutron:

Neutral

Found in Nucleus

Mass of 1 AMU

 

Electron:

Negative

Found in orbitals

Mass of 0 AMU

5 0
3 years ago
How many moles are in 12.9 grams of Fe?
amid [387]

Answer:

0.1677 is the answer

Explanation:

6 0
2 years ago
Read 2 more answers
ch sulfide (ZnS) occurs in the zinc blende crystal structure, (a) If 254 g of ZnS contains 170 g of Zn, what is the mass ratio o
Aleonysh [2.5K]

Answer:

The mass ratio of zinc to sulfide is 85:42.

2.5559 kg of Zn are in 3.82 kg of ZnS.

Explanation:

a) Mass of zinc sulfide = 254 g

Mass of zinc in a zinc sulfide sample = 170 g

Mass of sulfide in zinc sulfide sample = x

254 g = 170 g+ x

x = 84 g

The mass ratio of zinc to sulfide:

\frac{170 g}{84 g}=\frac{85}{42}

b) Mass of zincsulfide sample = 3.83 kg

The mass ratio of zinc to sulfide is 85:42.

Let the mass of zinc and sulfide be 85x and 42x respectively:

85 x+ 42 x=3.82 kg

x =0.03007 kg

Mass of an zinc= 85x=85 × 0.03007 kg= 2.5559 kg

8 0
3 years ago
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