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zmey [24]
2 years ago
15

A planet travels in an elliptical path around a star, as shown in the figure. As the planet gets closer to the star, the gravita

tional force that the star exerts on the planet increases. Which statement of reasoning best supports and correctly identifies what happens to the magnitude of the force that the planet exerts on the star as the planet gets closer to the star?
The force remains constant because the mass of the planet remains constant.
A

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.
B

The force decreases because the planet increases its speed as it gets closer to the star.
C

The force fluctuates such that it increases and decreases because the planet does not travel in a perfectly circular path.

Chemistry
1 answer:
Soloha48 [4]2 years ago
3 0

Answer:

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.

Explanation:

Force between two objects can be expressed by an equation:

F = G • m1 • m2 / r^2,

where m1 and m2 are objects' masses, r is the distance between them, and G is a gravitational constant.

That means that greater the masses or lesser the distance, the force will be greater, and vice versa.

This force exists between any two objects, but is generally extremely weak, so it's best observed with big and large objects with great mass, such as planets and stars.

This force, whatever its magnitude may be, always works on both objects, following the third Newton's law.

So, whatever the force the stat exerts on the planet is, the planet will exert the same amount of force on the star.

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Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of
NeX [460]

Answer:

Molarity of HCl = 0.19 M

Explanation:

Moles of Na_2CO_3:-

Mass = 0.246 g

Molar mass of Na_2CO_3 = 105.988 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.246\ g}{105.988\ g/mol}

Moles_{Na_2CO_3}= 0.002321\ mol

According to the given reaction:-

Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2

1 mole of Na_2CO_3 react with 2 moles of HCl

0.002321 mole of Na_2CO_3 react with 2*0.002321 moles of HCl

Moles of HCl = 0.004642 moles

Given that volume = 24.3 mL

Also,  

1\ mL=10^{-3}\ L

So, Volume = 24.3 / 1000 L = 0.0243 L

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.004642}{0.0243}

<u>Molarity of HCl = 0.19 M </u>

<u></u>

7 0
2 years ago
The compound Xe(CF3)2 decomposes in a first-order reaction to elemental Xe with a half-life of 30.0 min. If you place 4.5 mg of
GenaCL600 [577]

Answer : The time passed by the sample is, 1.2\times 10^2\text{ min}

Explanation :

Half-life = 30.0 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{30.0\text{ min}}

k=0.0231\text{ min}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.0231\text{ min}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 4.5 mg

a - x = amount left after decay process =0.25 mg

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0231}\log\frac{4.5}{0.25}

t=125.15\text{ min}=1.2\times 10^2\text{ min}

Therefore, the time passed by the sample is, 1.2\times 10^2\text{ min}

7 0
2 years ago
Commercially available hot packs are simple in design: a pouch with water on one side, isolated by a barrier from a specific sal
liq [111]

Answer:

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an exothermic reaction is when heat/light is produced. heart is produced from this reaction so it is exothermic

3 0
2 years ago
The standard hydrogen electrode:
tiny-mole [99]

Answer:

is a simple, safe, and easy-to-create electrode

Explanation:

the hydrogen electrode is based on the redox half-cell:

  • 2H+(aq)  +  2e- → H2(g)

building:

1) Platinium electrode

2) hydrogen pumping

3) acid solution [H+] = 1M

4) siphon to prevent oxygen presence

5) galvanic cell connector  

this electrode is used as the basis for standard potential tables

3 0
2 years ago
The recommended daily allowance (rda of calcium is 1.2 g. calcium carbonate contains 12.0% calcium by mass. how many grams of ca
defon
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!

2) Mass fraction of this is excessive data.

3) The solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
4 0
2 years ago
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