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Slav-nsk [51]
2 years ago
13

What does scientific inquiry refer to

Chemistry
1 answer:
Sloan [31]2 years ago
5 0
It refers to the ways scientists explore and study the world based on observations and their works/ evidence.
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Find the sum of angles x° and y° from the given figure​
tresset_1 [31]

Answer:

where is the figure . u can simply add them

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2 years ago
A hypothetical main group element E reacts with chlorine to form an ionic compound with the formula ECl. The element is a member
Ronch [10]
It would be 1A bc then the +1 charge will cancel out chlorine’s -1 charge
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3 years ago
Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering. B) invertebrates
tensa zangetsu [6.8K]

Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering.

The picture shows flowering plants that are using seeds, flowers or fruits and they are called flowering plant .

So the answer here is A) flowering.

Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering.

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Can somebody help me with this question please
Tasya [4]

Answer:

concave lens

Explanation:

it's concave lens because it diverges the ray/beam of light.

7 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
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