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2 years ago

What is an example in which a scientists might calculate the mean of a data set

1 answer:

3 0

An example is just when they have a a group o numbers, for any reason, and they want to find the average of then all

an example being um idk...numbers resulting from an experiment trying to find different densitys or something and they wanted to find the average density

You might be interested in

If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc

igor_vitrenko [27]

When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>

7 0

2 years ago

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0

2 years ago

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?

Xelga [282]

let the actual height of car be x

now, according to question,

$\dfrac{1}{15} = \dfrac{8}{x}$

$x = 15 \times 8$

$x = 120 \: \: cm$

$height \: \: of \: \: car = 120 \: cm \: \: or \: \: 1.2 \: \: m$

5 0

1 year ago

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

2 years ago

4. a student is shown a Diagram which shows gas bubbles being made by water plant that has been exposed to light. which gas is m

lisov135 [29]

Carbon dioxide is in the bubbles

5 0

2 years ago