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igor_vitrenko [27]
3 years ago
7

8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

8T is: (b) 3R
Chemistry
1 answer:
Elena-2011 [213]3 years ago
4 0

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

T^2\propto R^3

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3

(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3  

\dfrac{1}{64}=(\dfrac{R}{R'})^3

R'=4\times R

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.  

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What is the mole fraction of NaOH in an aqueous solution that
astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

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

Using the formula

Number of moles of  NaOH(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}

n_B=0.572moles

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water(n_A)=\frac{77.1}{18.02}

n_A=4.279 moles

Now, mole fraction of NaOH

=\frac{n_B}{n_B+n_A}

=\frac{4.279}{0.572+4.279}

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0
3 years ago
Using the periodic table, choose the more reactive nonmetal.<br> Te or O
jolli1 [7]

the answer is Te because O is oxygen.

5 0
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Please help<br>What is the possible ions of vanadium?
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V+5, V+4, V+3, and V+2

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Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene wi
Readme [11.4K]

Answer:

See explanation

Explanation:

When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.

The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.

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D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match th
Aleks04 [339]

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

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3 years ago
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