What do you think the scientific question is for the<br> boiling water experiment

Who was the first person to show strong empirical evidence for the existence of atoms?.

Mekhanik [1.2K]

Given what we know, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

<h3>Who was John Dalton?</h3>

He was a renowned scientist with knowledge in many fields.
He was known to be a chemist, meteorologist, and physicist.
He proposed the atomic theory and carried out experiments to provide supporting evidence.

Therefore, given his proposal of the atomic theory and the experiments he carried out to provide evidence to support his claims, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

To learn more about atoms visit:

brainly.com/question/13981855?referrer=searchResults

6 0

1 year ago

What was the biggest challenge you faced in getting to where you are today and how did you overcome it?

Alchen [17]

It’s all based on you but yours could be anything hard you went through and how you overcame it. Also say how it made you the person you are today

4 0

2 years ago

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33

Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

4 0

3 years ago

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate

vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^-(Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1= (244 + 273.15) K = 517.15 K

T2= (324+ 273.15) K =597.15 K

K1 = 6.7 , K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0

3 years ago

Read 2 more answers

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0

1 year ago

What do you think the scientific question is for the boiling water experiment￼

What do you think the scientific question is for the boiling water experiment