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Gemiola [76]
3 years ago
11

45 Three samples of the same solution are tested, each with a different indicator. All three indicators, bromthymol blue, bromcr

esol green, and thymol blue, appear blue if the pH of the solution is
(1) 4.7 (2) 6.0 (3) 7.8 (4) 9.9
Chemistry
2 answers:
zalisa [80]3 years ago
6 0
(4) 9.9 is the most sensible answer.

When an indicator appears blue, that means the solution is basic and the pH of the solution should be within the range of 8 and 14.

Hope this helps~


ANEK [815]3 years ago
3 0

<u>Answer:</u> The correct answer is Option 4.

<u>Explanation:</u>

Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.

The pH range and color change of these indicators are:

  1. Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
  2. Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
  3. Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6

As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.

Hence, the correct answer is Option 4.

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How many grams of oxygen gas will be produced when 2.50 moles of potassium chlorate is decomposed?
yawa3891 [41]

Answer:

m_{O_2}=120gO_2

Explanation:

Hello!

In this case, since the decomposition of potassium chlorate is:

2KClO_3\rightarrow 2KCl+3O_2

We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:

m_{O_2}=2.50molKClO_3*\frac{3molO_2}{2molKClO_3} *\frac{32.0gO_2}{1molO_2}\\\\m_{O_2}=120gO_2

Best regards!

8 0
3 years ago
N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

7 0
3 years ago
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Explanation:

Given:

Coulomb meter = -1 CM

Find:

In debye

Computation:

We know that,

1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye

So,

-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye

-1 Coulomb meter = -2.997 × 10²⁹ Debye

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