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2 years ago

What are some limitations of this experiment? How could it be improved?

Chemistry

2 answers:

ratelena [41]2 years ago

5 0

Answer:Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.

Explanation: i hope that helps

solong [7]2 years ago

3 0

Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.

Explanation:

Hope this helps!

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3 years ago

Which reflects more sunlight-dark soil or white sand?

Gnesinka [82]

I think the answer is white sand (sorry if i am wrong)

4 0

3 years ago

Read 2 more answers

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago

What does Einstein’s theory of relativity expalain?

anygoal [31]

Answer:

The correct answer is D. How gravity interacts with planets

Explanation:

Einstein's theory of relativity formulated in 1915 introduces the following concepts:

- Formula E = mc2 (relationship between energy, mass, speed of light)

- How nuclear energy is produced

-After seeing a solar eclipse, gravity is interpreted geometrically (which is an effect of distortion)

-Relative time

8 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas