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3 years ago

Which of the following is a reason why a scientist might do research?

Chemistry

1 answer:

timurjin [86]3 years ago

6 0

Answer:

A scientist does research to make sure he has the right answers to his problems and correct numbers.

Explanation:

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At a certain temperature Kc = 9.0 for the equilibrium 24() ⇔ 22(). What is

Iteru [2.4K]

Answer:

Explanation:

5 0

4 years ago

The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

7 0

3 years ago

Hey! I was wondering how to get the solution to this question!

zzz [600]

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

3 0

3 years ago

Why methylated spirits burn?

pishuonlain [190]

Answer:

$Methylated \: spirits \: are \: \: extremely \: flammable.$

Explanation:

Incorrect use has resulted in accidents and disfiguring burns. Never leave a methylated spirit appliance unattended. Also make sure that the camping stove or appliance is on a flat surface and that the fuel cannot spill out.

5 0

2 years ago

A chemist has some 40% acid solution, some 60% acid solution, and a wholebunch of free time. How many liters of each should be u

TiliK225 [7]

Answer:

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

Explanation:

Let volume of the 40% acid solution be x.

Let volume of the 60% acid solution be y.

Volume of solution formed after mixing both solution = 40 L

x + y = 40 L..[1]

Volume of acid 40% solution = 40% of x= 0.4x

Volume of acid 60% solution = 60% of y= 0.6y

Volume of acid formed = 45% of 40 L = $\frac{45}{100}\times 40=18L$

$0.4x+0.6y=18 L$ ..[2]

Solving [1] and [2]

x = 30 L , y = 10 L

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

8 0

3 years ago