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3 years ago

Why is it not possible to always have a controlled experiment

1 answer:

4 0

A controlled experiment is one in which evrerything is held constant except for one verieble, maybe is usually a st of data is taken for a control group.

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It is known that 32g of oxygen takes up 24 litres of space at 25oC. Calculate the volume of oxygen that is needed to react compl

Gwar [14]

6 L of oxygen (O₂)

Explanation:

We have the following chemical reaction:

2 Mg + O₂ → 2 MgO

Now we calculate the number of moles of magnesium:

number of moles = mass / molar weight

number of moles of Mg = 12 / 24 = 0.5 moles

Taking in account the chemical reaction we devise the following reasoning:

if 2 moles of Mg react with 1 mole of O₂

then 0.5 moles of Mg react with X moles of O₂

X = (0.5 × 1) / 2 = 0.25 mole of O₂

mass = number of moles × molar weight

mass of O₂ = 0.25 × 32 = 8 g

Knowing the information given by the problem that 32 g of O₂ takes up 24 liters of space, we devise the following reasoning:

if 32 g of O₂ takes up 24 liters of space

then 8 g of O₂ takes up Y liters of space

Y = (8 × 24) / 32 = 6 L of O₂

Learn more about:

balancing chemical reactions

brainly.com/question/14187466

#learnwithBrainly

4 0

3 years ago

You aren't sure if your pH meter needs to be calibrated, so you put it in a buffer solution that has a known pH of 7. You take f

Roman55 [17]

Answer:

Neither

Explanation:

Even though the average between the 4 measurements is 7.45, the measurements are not accurate. This is probably due to the pH meter being out of calibration or because there is some contamination in the buffer solution

5 0

4 years ago

A quantity of gas weighing 7.1 g at 741 torr and 44 celius occupies 5.41 L. Find its molar mass.

Aleks04 [339]

1 atmosphere = 760 torr
R = 0.082
44 degrees Celsius = 317 degrees Kelvin

$PV=NRT\\ \frac{744}{760}(5.41)=N(0.082)(317)\\ \frac{4025.04}{760}=25.994N\\\frac{4025.04}{19755.44}=N\\N=\frac{50313}{246943}\\0.203743$

3 0

3 years ago

What is the molar volume of a gas at standard temperature and pressure?

Andrej [43]

Answer: The molar volume of any gas at standard pressure and standard temperature is 22.4 liters per mole.

Explanation:The ideal gas law is PV=nRT

P is pressure and if we consider standard pressure, then we have 1.00 atm.

V is volume and that is what we are trying to solve.

n is the number of moles, which is 1.00 moles since we are trying to determine the volume of a gas in one mole.

R is the ideal gas constant which equals

0.0821 (Liters x atmospheres)/(mole x kelvin)

T is the standard temperature which is 273 kelvin.

Rearrange the equation to solve for volume.

V = nRT/P

V = (1.00 mol)(0.0821 L atm/mol K)(273 K)/ 1.00 atm

V = 22.4 L

4 0

3 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

lana [24]

Answer:

Explanation:

Given data:

Mass of calcium oxide = 14.4 g

Mass of carbon dioxide = 13.8 g

Actual yield of calcium carbonate = 19.4 g

Mass of calcium carbonate produced = ?

Limiting reactant = ?

Percent yield = ?

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles of CaO = Mass /molar mass

Number of moles of CaO = 14.4 g / 56.1g/mol

Number of moles of CaO = 0.26 mol

Number of moles of CO₂:

Number of moles of CO₂= Mass /molar mass

Number of moles of CO₂ = 13.8 g / 44 g/mol

Number of moles of CO₂ = 0.31 mol

Now we will compare the moles of CaCO₃ with CO₂ and CaO.

CaO : CaCO₃

1 : 1

0.26 : 0.26

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Limiting reactant:

CaO

Theoretical yield:

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ = 0.26 mol × 100 g/mol

Mass of CaCO₃ = 26 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 19.4 g/ 26 g× 100

Percent yield = 74.6 %

3 0

3 years ago