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maria [59]
3 years ago
5

Explain the importance of replication in scientific research

Chemistry
2 answers:
SVEN [57.7K]3 years ago
8 0

Answer:

Explanation:

Replication is crucial to science. Every time researcher find something surprising that has to do with science they replicate it.

meriva3 years ago
4 0
Rawr grrr rawr grrr rawr grr
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Answer all questions pls
tekilochka [14]

Answer:

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4) Temperature, humidity, precipitation, air pressure, wind speed, and wind direction are key observations of the atmosphere that help forecasters predict the weather. These same factors have been used since the first weather observations were recorded. Observational data collected by doppler radar, radiosondes, weather satellites, buoys and other instruments are fed into computerized NWS numerical forecast models. The models use equations, along with new and past weather data, to provide forecast guidance to our meteorologists. The three main factors of weather are light (solar radiation), water (moisture) and temperature.

Explanation:

U dont have to copy and paste this put these are some ideas to use for ur answers

6 0
2 years ago
Question 2
san4es73 [151]

Explanation:

its hard to explain unless we know what the question fully asks..

4 0
2 years ago
Noble gas notation for platinum
SOVA2 [1]

Pt

Hope this helps!!!

4 0
3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
2 years ago
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