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Korolek [52]
3 years ago
12

What’s a synonym for control group

Chemistry
1 answer:
viktelen [127]3 years ago
3 0

Answer:

comparison group

reference group

comparator group

pilot group

monitoring group

Explanation: you could use one or another

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tensa zangetsu [6.8K]

meaning:An acid is a chemical substance, usually a liquid, which contains hydrogen and can react with other substances to form salts. Some acids burn or dissolve other substances that they come into contact with.

5 0
3 years ago
Read 2 more answers
Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O
Nikolay [14]

Answer:

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

<u>1 mole of substance</u> = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

<em>   1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles</em>

Converting grams to moles:

<em>Given grams => moles = grams/gram formula wt </em>

Converting moles to grams:

<em>Given moles => grams = moles x gram formula wt</em>

_________________________________________________________

<em>Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.</em>

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass <em>CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O</em>

<em>= 0.364 g CaSO₄·2H₂O </em><em>/ </em><em>172 g CaSO4·2H2O </em>

<em>= (0.364/172) mole CaSO₄·2H₂O </em>

<em>= 2.12 x 10⁻³ mole CaSO₄·2H₂O    </em>

<em>∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O </em>

<em>=  (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)</em>

<em>= 1.276876 x 10²¹molecules CaSO₄· 2H₂O  CaSO₄2H₂O </em>

<em>= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O   x   6 oxygen atoms / molecule</em>

<em>= 7.661256 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em />

8 0
3 years ago
Based on the Hertzprung-Russell diagram, which statement is true? OA. Main sequence stars with low temperatures tend to have low
ololo11 [35]

Main sequence stars with low temperatures tend to have low luminosity is based on Hertzprung-Russell diagram

<h3>What is the Hertzsprung-Russell diagram's main sequence?</h3>

The bright, hot stars in the image's upper left are the main sequence group, which stretches roughly diagonally to the lower right (dim and cool). In the bottom left are white dwarfs, which are small, faint, and hot. In the top right are giant and supergiant stars, which are big, bright, and cool.

<h3>What is the main sequence star H-R diagram trend?</h3>

Main sequence stars have a wide range of effective temperatures, but because they are brighter at higher temperatures, they normally follow a band from the bottom right to the top left of the diagram. Inside the cores, hydrogen and helium are fusing together.

To know more about H-R diagram visit:

brainly.com/question/13635125

#SPJ9

6 0
2 years ago
Read 2 more answers
Suppose you wanted to cause Pb2+ ions to come out of solution as solid Pb. Which metal(s) could you use to accomplish this? Au Z
lions [1.4K]

Answer:

Ca, Zn, Ni, Al, Fe

Explanation:

The activity series is an arrangement of metals in order of decreasing reactivity.

Metals that are high up in the activity series are very reactive. They displace metals that are lower in the activity series from an aqueous solution of their salts.

The metals; Ca, Zn, Ni, Al, Fe are all above Pb in the activity series hence they will cause Pb2+ ions to come out of solution as solid Pb.

5 0
3 years ago
A compound contains 34.5% calcium, 24.1% silicon and 41.4% oxygen by mass. What is its empirical formula?
Amiraneli [1.4K]

<em>empirical \: formula \\  = CaSiO3 \\ please \: see \: the \: attached \: picture \\ hope \: it \: helps</em>

4 0
3 years ago
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