Thomson<span> is the scientist who designed an experiment that enabled the first successful detection of an individual subatomic particle. </span>J.J. Thomson<span> (Sir </span>Joseph John Thomson<span>, 1856-1940), who demonstrated in 1897 that "cathode rays" consisted of negatively-charged particles, later named electrons.</span>

6 0

3 years ago

How many mL of a 15% acid solution and how many mL of a 3% acid solution must be mixed to get 45 mL of a 7% acid solution?

madam [21]

Answer:

15 mL of 15% acid solution and 30 mL of 3% acid solution

Explanation:

The total number of moles of the mixture $(n_{3})$ is equivalent to the addition of the number of moles of the first solution $(n_{1})$ and the number of moles of the second solution $(n_{2})$ . Mathematically,

$n_{1} +n_{2} = n_{3}$

$n_{1} = V_{1}*C_{1}$

$n_{2} = V_{2}*C_{2}$

$n_{3} = V_{3}*C_{3}$

The volume of the mixture = 45 mL

The volume of first solution = x

The volume of second solution = 45 - x

Therefore:

0.15x + 0.03(45-x) = 0.07*45

0.15x + 1.35-0.03x = 3.15

0.12x = 1.8

x = 15

Thus, the volume of the first solution is 15 mL while the volume of the second solution is 30 mL.

3 0

3 years ago

A 22.4g sample of a substance was added to a graduated cylinder. It caused an 18.3 mL change in the volume of the water in the c

11111nata11111 [884]

<h3>Answer:</h3>

Density = 1.22 g.mL⁻¹

<h3>Solution:</h3>

Data Given:

Mass = 22.4 g

Volume = 18.3 mL

Density = ??

Formula used;

Density = Mass ÷ Volume

Putting values,

Density = 22.4 g ÷ 18.3 mL

Density = 1.22 g.mL⁻¹

5 0

3 years ago

Consider a sample of 10.0 g of the gaseous hydrocarbon C2H6 to answer the following question: How many moles are present in this

Masja [62]

<u>Answer:</u> The moles of given hydrocarbon is 0.3 moles

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of ethane = 10.0 g

Molar mass of ethane = $[(2\times 12)+(6\times 1)]=30g/mol$

We need to divide the given value by the molar mass.

Putting values in above equation, we get:

$\text{Moles of ethane}=\frac{10.0g}{30g/mol}=0.3mol$

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits. Here, the least precise number of significant digits are 1.

Hence, the moles of given hydrocarbon is 0.3 moles

7 0

3 years ago