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Lemur [1.5K]
3 years ago
12

What are the examples of all the non metals in the first twenty element​

Chemistry
1 answer:
agasfer [191]3 years ago
5 0

Answer:

semimetals or metalloids.

Explanation:

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HELP PLS 50 POINTSSSS
Alenkasestr [34]

Answer:

Carbon dioxide is a product (cellular respiration)

Carbon dioxide is a reactant(photosynthesis)

Carried out in animals(cellular respiration)

Carried out in plants(both)

Chemical reaction(cellular respiration)

Oxygen is a product(photosynthesis)

Oxygen is a reactant(cellular respiration)

Produces usable energy source(photosynthesis)

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4 0
3 years ago
Read 2 more answers
given a mixture of lead to oxide ammonium chloride and sodium chloride describe how a mixture can be separated to obtain a sampl
wariber [46]

Answer:

There are way in separating mixtures of chloride salts such as that of sodium chloride and ammonium chloride. It can be done by crystallization, filtration, and sublimation. The easiest method among the three is separation by sublimation.

5 0
3 years ago
What are some limitations of this experiment? How could it be improved?
solong [7]

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!

3 0
2 years ago
Read 2 more answers
How are electrons arranged around an atom?
Schach [20]
They are arranged in shells
6 0
2 years ago
How much ch2o is needed to prepare 445 ml of a 2.65 m solution of ch2o?
worty [1.4K]

Answer: 35.4 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

Molality = 2.65

n= moles of solute =?

 V_s = volume of solution in ml = 445 ml

Putting in the values we get:

2.65=\frac{n\times 1000}{445ml}

n=1.18

Mass of solute in g=moles\times {\text {molar mass}}=1.18mol\times 30.02g/mol=35.4g

Thus 35.4 grams of CH_2O is needed to prepare 445 ml of a 2.65 m solution of CH_2O.

8 0
2 years ago
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