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2 years ago

Aristarchus tried to calculate the sizes of the sun and moon based on

1 answer:

8 0

The incorrect statement is that; "Aristarchus calculated the distances between the sun and earth fairly accurately."

<h3>The size of the solar system</h3>

The solar system is composed of the sun, the planets and other celestial bodies such as the moon. We can see form the experiment that Aristarchus tried to determine the relative sizes of the earth and the moon.

However, the incorrect statement is that; "Aristarchus calculated the distances between the sun and earth fairly accurately."

Learn more about the solar system: brainly.com/question/12075871

You might be interested in

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

4 years ago

how does the total mass of each object increases the amount of force that is needed to get them moving at 5 m/s increase by abou

irakobra [83]

Answer:

The answer is below

Explanation:

Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum with respect to time, going in the same direction as the force.

Let F = force, m = mass of object, v = velocity of object, mv = momentum.

F = d/dt(mv) = m(dv / dt) = ma; a = acceleration.

Let us assume that the object starts from rest to 5 m/s within 1 seconds, hence:

F = m(dv / dt)

200 N = m[(5 m/s - 0 m/s) / (1 s)]

200 = 5m

m = 40 kg

7 0

3 years ago

How many grarns would 3 moles of carbon dioxide (CO2) weigh? (round to the

BigorU [14]

Mass of CO₂ = 132 g

<h3>Further explanation </h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mole also can be formulated :

$\tt n=\dfrac{mass}{MW}$

moles of CO₂ = 3

mass of CO₂(MW=44.01 g/mol) :

$\tt mass=mol\times MW\\\\mass=3\times 44.01\\\\mass=132.03~g\approx 132~g$

6 0

3 years ago

Michelle learns in science class that simple machines such as an inclined plane can change the amount of force needed to lift he

storchak [24]

Answer: the amount of force on the spring scale

Explanation:

4 0

3 years ago

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0

3 years ago