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1 year ago

Which of the following air pollutants is correctly paired with one of its major effects?

Chemistry

1 answer:

Alik [6]1 year ago

3 0

Answer:

The correct option is;

Sulfur oxides - acid precipitation

Explanation:

Here we have sulfur oxide when present in the atmosphere and mixed with oxygen water and other chemicals form acidic precipitation known as acid rain.

The sulfur oxides reacts with water in the clouds to form sulfuric acid as follows;

The sulfur gas is first oxidized

SO₂ + OH → HOSO₂

The next stage is the formation of sulfur trioxide

HOSO₂ + O₂ → HO₂ + SO₃

Sulfur trioxide combines with water to form sulfuric acid

SO₃ + H₂O → H₂SO₄ (aq).

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The temperature on a distant, undiscovered planet is expressed in degrees B. For example, water boils at 180 ∘ B and freezes at

marin [14]

Answer:

40.3∘C

Explanation:

At planet B;

Water boils = 180∘C

Water freezes = 50∘C

In this planet the temperature difference = 180 - 50 = 130 compared to earth where the temperature difference is; 100 - 0 = 100

This means;

130 ∘C = 100 ∘C

x ∘C = 31 ∘C

x = 31 * 130 / 100

x = 40.3∘C

5 0

2 years ago

In a closed system, how will a decrease in pressure affect the following reaction: 2A(g) +2B(g) ⇌ 2C(g) + 2D(g)?

DochEvi [55]

As number of gaseous moles in reactant and prodict are same that is 4

So No change will occur

3 0

2 years ago

Read 2 more answers

Which compound forms a green aqueous solution

Irina18 [472]

NiCl₂ commonly forms a green aqueous solution.

3 0

2 years ago

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

1 year ago

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