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Setler79 [48]
3 years ago
5

Water and carbon dioxide are similar because they both contain at least one oxygen

Chemistry
1 answer:
nadezda [96]3 years ago
6 0
Yes that is true.
Water: H_2 O
Carbon dioxide: CO_2
Both have O
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What does 10% humidity in a weather report indicate?
GalinKa [24]

Answer:

B

the humidity is the % of water in the air and it is out of 100% of the air. so 10/100 is low

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Does methylenecyclohexane obey Zaitsev's rule? why or why not
kifflom [539]

Answer:

Explanation: Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via E1 mechanism or via E2 mechanism. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, i.e., the alkene with the more highly substituted double bond.

E1 reaction always follow Zaitsev’s rule; with E2 reactions, there are exceptions (see antiperiplanar).

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A mixture of uneven distribution and easy separation is called_____. homogeneous heterogeneous a solution a pure substance
dem82 [27]
The correct answer is heterogeneous.
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36g of an alloy of copper and zinc contains 45% copper. how much pure copper do you need to add in order to get a 60% copper all
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First, you need to count copper mass in alloy.
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7 0
3 years ago
A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.
adell [148]

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

c) \rho=1.5x10^5g/L

Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

8 0
4 years ago
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