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3 years ago

Students in chemistry class have been given the assignment to use flame test emission data to determine the identity of an

Chemistry

1 answer:

konstantin123 [22]3 years ago

6 0

It is impossible to narrow down anything about the unknown.

Flame tests are common analytical techniques used to determine the composition of an unknown compound based on the color that it imparts to flame. Each metal ion imparts a specific color to flame which helps us to identify it.

In this case, two colors were visible in the flame and it will be difficult to determine which of the colors corresponds to the actual metal ion present.

Therefore, it is impossible to narrow down anything about the unknown.

Learn more: brainly.com/question/6357832

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What is the difference between an amoeba and a hydra?

Marat540 [252]

An amoeba is a single felled organism

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3 years ago

How many C atoms are there in a sample of C,H, that contains 6.59 x 1026 H atoms?

Flauer [41]

Answer:

bill gates

Explanation:

4 0

3 years ago

A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.Part A: What is the initial pH of this solution?Part B: What

Lelechka [254]

Answer:

Initial pH of this pH = 7.453

pH after addition of 150.0 mg of HBr = 7.35

pH after addition of 85.0 mg of NaOH 0.154 = 7.56

Explanation:

Since Ka value isn't given

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log $\frac{[Conjugate base]}{[Acid]}$

pH = 7.52 + log $\frac{0.15}{0.175}$

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr

$\frac{Mass}{Molar mass} \\= \frac{150 X10^{-3}g }{80.91} \\= 0.00185 } mole$

Moles of NaOCl in 100 ml buffer solution = $\frac{0.15X100}{1000}$ = 0.015

Moles of HClO in 100 ml buffer solution = $\frac{0.175X100}{1000}$ = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

= 0.0132

moles of HClO = 0.0175 + 0.00185

= 0.0194

Using Henderson equation pH = Pka + log $\frac{Conjugate base}{Acid}$

= 7.52 + log $\frac{0.0132}{0.0194}$

= 7.35

Part C

pH after addition of 85 mg of HBr

moles of NaOH

$\frac{Mass}{Molar mass} \\= \frac{85 X10^{-3}g }{40} \\= 0.00213 } mole$

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

= 0.0171

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

= 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log $\frac{Conjugate base}{Acid}$

= 7.52 + log $\frac{0.171}{0.154}$

= 7.56

3 0

3 years ago

Which of the following statements is true? (Multiple answers possible) Group of answer choices:

Sophie [7]

Answer:The endpoint does not correspond exactly to the equivalence point

At the endpoint, a change in a physical quantity associated with the equivalence point occurs.

At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present.

Explanation:

The end point is always indicated by some physical property that changes such as colour. At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present. The equivalence point cannot be physically observed but can be deduced after a titration curve is plotted.

3 0

4 years ago

What pattern do you observe in the coefficients of the hydrocarbon (CxHy) and water (H2O)? How do you explain this pattern?

balu736 [363]

The coefficients of the hydrocarbon and water is 61 and it is explained below.

Explanation:

We have to balance the chemical equation that describes the combustion of octane C₈H₁₈. The combustion of octane involves burning this hydrocarbon in the presence of excess oxygen,O 2. Because octane is a hydrocarbon, that is the compound that contains only carbon and hydrogen, the reaction will produce two products carbon dioxide CO 2 and water H 2 O.

The unbalanced chemical equation is

C₈H₁₈ + O₂ → CO₂ + H₂O

To balance this equation, we have 8 present on the reactant side, so multiply the carbon dioxide by 8 to get 8 atoms of carbon on the products side.

C₈H₁₈ + O₂ → 8 CO₂ + H₂O

Now, we have 18 on the reactant side and 2 on the products side, so multiply the water molecule by 9 to get

C₈H₁₈ + O₂ → 8 CO₂ + 9 H₂O

8 * 2 atoms O + 9 *1 atoms O = 25 atoms O

By adding fractional coefficient to O₂ we get

C₈H₁₈ + $\frac{25}{2}$ O₂ → 8 CO₂ + 9 H₂O

Multiply all the coefficients by 2 we get,

2C₈H₁₈ + ( $\frac{25}{2}$ * 2) O₂ → 8 * 2 CO₂ + 9* 2 H₂O

The balance equation is

2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

Add the coefficients to get their sum

= 2 +25 + 16 +18 = 61

Thus, The coefficients of the hydrocarbon and water is 61

6 0

3 years ago