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3 years ago

Which of the different components of smog depicted in the graph are most likely released from automobile exhaust?

Chemistry

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

B, B and C

Explanation:

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Chemistry!! Please help

m_a_m_a [10]

Answer:

5.3 × 10⁻¹⁷ mol·L⁻¹

Explanation:

Let s = the molar solubility.

Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq); K_{sp} = 6.1 × 10⁻⁴⁹

E/mol·L⁻¹: 2s s

K_{sp} =[Cu⁺]²[S²⁻] = (2s)²×s = 4s^3 = 6.1 × 10⁻⁴⁹

$s^{3}= \frac{6.1 \times 10^{-49}}{4} = 1.52 \times 10^{-49}$

$s = \sqrt[3]{1.52 \times 10^{-49}} \text{ mol/L} = 5.3 \times 10^{-17} \text{ mol/L}$

6 0

3 years ago

In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole pro

Ksju [112]

Based on experiment 1:

Mass of Hg = 1.00 g

Mass of sulfide = 1.16 g

Mass of sulfur = 1.16 - 1.00 = 0.16 g

# moles of Hg = 1 g/200 gmol-1 = 0.005 moles

# moles of S = 0.16/32 gmol-1 = 0.005 moles

The Hg:S ratio is 1:1, hence the sulfide is HgS

Based on experiment 2:

Mass of Hg taken = 1.56 g

# moles of Hg = 1.56/200 = 0.0078

Mass of S taken = 1.02 g

# moles of S = 1.02/32 = 0.0319

Hence the limiting reagent is Hg

# moles of Hg reacted = # moles of HgS formed = 0.0078 moles

Molar mass of HgS = 232 g/mol

Therefore, mass of HgS formed = 0.0078 * 232 = 1.809 g = 1.81 g

3 0

3 years ago

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M ac

adell [148]

Explanation:

Below is an attachment containing the solution.

7 0

4 years ago

How many atoms are in 1.4 mol of phosphorus trifluoride (PF3)?

joja [24]

Number of atoms in 1.4 mol of Phosphorus trifluoride (PF₃) : 8.428 x 10²³

<h3>Further explanation </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

1.4 mol of Phosphorus trifluoride (PF₃), number of atoms :

$\tt N=1.4\times 6.02\times 10^{23}\\\\N=8.428\times 10^{23}$

5 0

3 years ago

Uranium-236 is _____. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-

nata0808 [166]

Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.

Following is the reaction involved in above process:

 $^{235}U$ + $^{1}n$ → $^{236}U$ → $^{144}Ba$ + $^{89}Kr$ + 3 $^{1}n$ + 177 MeV

Here,
$^{235}U$ = Fission material,
$^{1}n$ = projectile,
$^{236}U$ = intermediate nuclei,
$^{144}Ba$ and $^{89}Kr$ = Fission product

8 0

3 years ago

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