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Setler [38]
2 years ago
12

Mass is a measure of weight. A. True B. False

Chemistry
2 answers:
Cerrena [4.2K]2 years ago
6 0

Answer:

A. True

Explanation:

notka56 [123]2 years ago
4 0

Answer: haii~! ur answer is A.) true

Explanation: <u><em>hope dis helps ^^</em></u>

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Can you help me answer this plz
Zigmanuir [339]

Answer: In Galileo’s time, what was considered the “center of everything”? The Earth! All of the planets and even the Sun went around “us”. Of course, when Galileo saw the moons of Jupiter passing in front of the planet, and disappearing to show up again, it was clear to him that these moons went around Jupiter like our moon goes ‘round the Earth.

That did not set well with the Beliefs of the day, and that is at least one answer!

Explanation:

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What are Free Radicals ?​
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Answer:

an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.

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"three points A, B, and C are collinear, and B is between A and C, then AB + BC = AC” is a statement of the ___ postulate
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Segment addition postulate.
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3 years ago
This pair of molecules are _____.The figure shows two ball-and-stick models of molecules. The first consists of five black balls
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Alkanes

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The description of the first molecule corresponds to a cycloalkane. In the cycloalkanes for each carbon atom there are 2 hydrogen atoms (CnH2n = C5H10).

The second molecule corresponds to an open chain alkane, where for every carbon atom there is double plus 2 hydrogen atoms (CnH (2n + 2) = C5H12)

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3 years ago
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For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

4 0
3 years ago
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