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dybincka [34]
4 years ago
12

Sulfur dioxide gas, SO2(g), is released from active volcanoes. It reacts in the atmosphere to form sulfur trioxide, SO3(g).

Chemistry
2 answers:
aksik [14]4 years ago
3 0

Explanation:

The reaction given:

2SO2(g) + O2(g) -> 2SO3(g)

A. "The reaction is balanced."

We multiply the sulphur atoms on both sides:

2S -> 2S, ......... 2 sulphur atoms on each side, so balanced.

now oxygen:

2O2+O2 -> 2O3........6 atoms on each side, so also balanced.

This means that the equation is balanced.

B. "When 2 L SO2 react with excess O2, 2 mol SO3 form "

The coefficient of reactant SO2 is 2, so is that of the product.  If we had the same units on either side of the equation (mols, litres, molecules), the statement would have been true.  Since 2L  of SO2 does not contain 2 mol of SO2, so the statement is no longer true.

C. "The coefficients can represent moles, molecules, or liters"

As mentioned in (B) above, we can use moles, molecules or litres (or any other volume unit) to apply to the coefficient and the equation will still be true.  So the statement is true.

Novosadov [1.4K]4 years ago
3 0

Answer:

1st and 3rd options

Explanation:

got it right on edu

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What contribution did Robert Boyle make towards the field of chemistry
GREYUIT [131]

Answer:

Boyle's law, which describes the inversely proportional relationship between the absolute pressure and volume of a gas, if the temperature is kept constant within a closed system. Among his works, The Sceptical Chymist is seen as a cornerstone book in the field of chemistry.

8 0
3 years ago
KOH + HBr - KBr + H2O<br> Which is the acid in this reaction?
Reptile [31]

Answer:

HBr is a strong acid

Explanation:

KBr is a salt which makes a base . also KOH is a base

7 0
4 years ago
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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
3 years ago
5. Write the chemical formulae of the following chemical compounds.
Vladimir [108]

Answer:

  1. Na2SO4
  2. KNO3
  3. Cu2O.
  4. H2SO4.
  5. ZnSO
  6. (AlCl3)
  7. Ca3N2
  8. H²S
  9. Fe2O3
  10. MnO²

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4 0
3 years ago
Lucite contains 59.9 g C, 8.06 g H,
Alexxandr [17]
You start by finding the mol of each

59.9g C x (mol C / 12.01 g C) = 4.98 mol C
8.06g H x (mol H / 1.00 g H) = 8.06 mol H
32.0 g O x (mol O / 16.0 g O) = 2 mol O

So when you set it up you have
C4.98 H8 O2

You divide each by the smallest mol. The smallest mol is 2

C2 H4 O2.5

However you can’t have half a mol in the empirical formula. If the value ends in 0.5, you multiply everything by 2

You’re left with

C2H8O5

The EMPIRICAL formula for lucite is C2H8O5

Note empirical is not the same as chemical formula.
6 0
2 years ago
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