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3 years ago

5

BALANCE THAI CHEMICAL EQUATION PLS, help:)

1 answer:

Zinaida [17]3 years ago

8 0

Answer:

4C₈H₄OH + 35O₂ → 32CO₂ + 10H₂O

Explanation:

The unbalanced reaction expression is given as:

C₈H₄OH + O₂ → CO₂ + H₂O

To solve this problem, we use a mathematical approach.

aC₈H₄OH + bO₂ → cCO₂ + dH₂O

Conserving C: 8a = c

H: 5a = 2d

O: a + 2b = 2c + d

Let a = 1, c = 8, d = $\frac{5}{2}$ , b = 35

4C₈H₄OH + 35O₂ → 32CO₂ + 10H₂O

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Give the formula for the compound formed when sulfur trioxide reacts with water.

Lera25 [3.4K]

Sulphur trioxide reacts with water to form a solution of sulphuric acid.
The equation describing this reaction is:
SO3 + H2O .............> H2SO4

Based on the above equation, the formula of the compound formed when sulphur trioxide reacts with water is: H2SO4

6 0

3 years ago

At 2000 ∘c the equilibrium constant for the reaction 2no(g)⇌n2(g)+o2(g) is kc=2.4×103. part a if the initial concentration of no

luda_lava [24]

Answer:

[NO] = 1.72 x 10⁻³ M.

Explanation:

For the reaction:

<em>2NO(g) ⇌ N₂(g)+O₂(g),</em>

Kc = [N₂][O₂] / [NO]².

At initial time: [NO] = 0.171 M, [N₂] = [O₂] = 0.0 M.

At equilibrium: [NO] = 0.171 M - 2x , [N₂] = [O₂] = x M.

∵ Kc = [N₂][O₂] / [NO]².

∴ 2400 = x² / (0.171 - 2x)² .

<u><em>Taking the aquare root for both sides:</em></u>

√(2400) = x / (0.171 - 2x)

48.99 = x / (0.171 - 2x)

48.99 (0.171 - 2x) = x

8.377 - 97.98 x = x

8.377 = 98.98 x.

∴ x = 8.464 x 10⁻².

<em>∴ [NO] = 0.171 - 2(8.464 x 10⁻²) = 1.72 x 10⁻³ M. </em>

<em>∴ [N₂] = [O₂] = x = 8.464 x 10⁻² M.</em>

3 0

3 years ago

What are the two parts of a solution?

Vlad [161]

The 2 parts or components that make up a solution would be the solute and the solvent.

6 0

3 years ago

Liquid butane, C4H10, is stored in cylinders to be used

Ganezh [65]

Answer:

<u>13.0kJ</u>

Explanation:

The<em> heat</em> to <em>vaporize</em> a l<em>iquid</em> is equal to the amount of liquid in moles multiplied by the specific <em>heat of vaporiztion</em> per mole.

First, calculate the number of moles in 35.5g of <em>butane</em>.

Molar mass of butane: 58.124 g/mol

Number of moles = mass in grams/molar mass

Number of moles = 35.5g / 58.124g/mol = 0.6107632mol

Now, calculate the heat to vaporize that amount of <em>liquid butane</em>:

Heat = number of moles × specific heat of vaporization

Heat = 0.6107632mol × 21.3kJ/mol = 13.0 kJ

The answer must be reported with 3 significant figures.

3 0

4 years ago

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximat

trapecia [35]

Answer:

4.76

Explanation:

In this case, we have to start with the <u>buffer system</u>:

$CH_3COOH~->~CH_3COO^-~+~H^+$

We have an acid ( $CH_3COOH$ ) and a base ( $CH_3COO^-$ ). Therefore we can write the <u>henderson-hasselbach reaction</u>:

$pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}$

If we want to calculate the pH, we have to <u>calculate the pKa</u>:

$pH=-Log~Ka=4.76$

According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:

$[CH_3COO^-]=[CH_3COOH]$

If we divide:

$\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1$

If we do the Log of 1:

$Log~1=~zero$

So:

$pH~=~pKa$

With this in mind, the pH is 4.76.

I hope it helps!

8 0

3 years ago