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inessss [21]
3 years ago
13

Which change in the sun’s gravity would occur if the sun’s mass were reduced by one-half?(1 point)

Chemistry
2 answers:
Kisachek [45]3 years ago
8 0

Answer:

Gravity or the gravitational force is defined as the force of attraction occurring between all states of matter. The sun’s gravity, mass, and radius plays a crucial role in the properties of the Sun.

Explanation:

The gravity of an object can be determined using the formula:

\text{g}&=\frac{\text{GM}}{\text{R}^{2} }}

Now, according to the question, the mass of the sun is halved, which will be written as:

\text{g}&=\frac{\text{G}\frac{\text{M}}{2}}{\text{R}^{2} }}

Thus, according to the equation, the mass of an object is directly proportional to the gravitational field. since, the mass of the Sun is reduced to half, the gravity of the Sun will be halved.

Rudik [331]3 years ago
8 0

To find what will change in the sun's gravity, we will make use of the earth's gravity equation.

<u><em>B: It would be halved</em></u>

From the gravity equation, we know that the gravity is gotten from the formula;

g = GM/R²

where;

M is the mass of the Earth,

R the radius of the Earth

G is the gravitational constant = 6.67 × 10⁻¹¹ m³/kg.s²

From the gravity equation above, we see that the gravity is directly proportional to the mass.

Thus, if the mass was reduced by one - half, it means that the gravity will also be reduced by one-half.

Thus, we can conclude that the gravity of the Sun will be halved.

Read more at; brainly.in/question/5641471

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300

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at least 300 molecules

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Why is the temperature needed to freeze ocean water lower than the temperature needed to freeze the surface of a freshwater lake
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Zarrin [17]

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8 0
2 years ago
g modenr vacuum pumps make it easy to attain pressures of the order of in the laboratory. at a preasusure of 6.75 atm and an ord
Ratling [72]

Answer:

Number of molecules = 1.8267×10^20

Explanation:

From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;

PV = nRT

where

P =pressure

V =volume

n = the number of moles

R is the gas constant equal to 0.0821 L·atm/mol·K

T is the absolute temperature

Given:

P = 6.75 atm;

T = 290.0 k,

; V = 1.07 cm³ = 0.001 L

( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)

n = 3.0335167*10^-4 moles

But there are 6.022×10²³ molecules in 1 mole,

Number of molecules = 1.8267×10^20

7 0
3 years ago
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
3 years ago
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