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lisov135 [29]
3 years ago
13

Drag force is directly proportional to ______. velocity. thrust shorter and faster blades longer and slower blades

Chemistry
1 answer:
aniked [119]3 years ago
3 0

Answer:

Drag force is directly proportional to squared velocity

Explanation:

googl e

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Develop a message for the owners of cars in villages explaining why the vehicles should be kept well even if they are spoiled, w
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<h2>Answer:</h2>

<h3>#Carry On Learning</h3>

<h2>Explanation:</h2>

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Which of the following planets has ammonia clouds in its atmosphere?
AlekseyPX

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Jupiter and Saturn has ammonia clouds in its atmosphere since the molecules of ammonia gas that are present in their atmospheres condense to form clouds. However, these ammonia clouds are more visible in Jupiter than in Saturn because of their lower altitude in Saturn.

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PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

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3 years ago
Any groups present on a benzene ring can impact the success and regioselectivity of an electrophilic aromatic substitution. Dete
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Answer:

Explanation:

In electrophilic aromatic substitution, the substituent(s) that favors electrophilic attack is said to be ortho or para to the substituent. Meta directors are deactivators. Altogether, these directors have an integral role to play in the success and regioselectivity of electrophilic aromatic substitution reactions.

From the information given: The;

Strongly deactivating meta- director  ⇒ NO₂ (nitro group)

Moderately deactivating meta- director  ⇒ CN (cyano group)

Strongly activating ortho-/para- director ⇒ OH (hydroxy; group)

Weakly deactivating ortho-/para- director ⇒ -X (halide group)

Weakly activating ortho-/para- director ⇒ - Ph  (phenyl group)

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