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3 years ago

The equal areas law is Keplers 2nd law of motion of planetary motion. It states that _____

2 answers:

7 0

Answer: It states that a line joining a planet and the sun sweeps an equal area in equal time, moving along its orbit.

Explanation: It tells us about the speed of a planet revolving around the sun in an elliptical orbit. The speed of the planet varies inversely to the distance, when the planet is closer speed is high and speed is low when it recedes from the sun. The rate at which area is being swept out is constant ,that is 'equal area in equal time'.

seraphim [82]3 years ago

6 0

An imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time. Thus, the speed of the planet increases as it nears the sun and decreases as it recedes from the sun.

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The answer to a chemistry question

Otrada [13]

<span>4 C3H5(NO3)3 = 12 CO2 + 6 N2 + 10 H2O + O<span>2
That's the answer lovely~</span></span>

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3 years ago

Iupac names of the following

Nadusha1986 [10]

It’s blurry.... try again

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2 years ago

Which type of wave do the particles in a medium move parallel to the direction that the waves move

Talja [164]

Longitudinal waves. In a longitudinal wave the particles in the medium move parallel to the direction waves go. A good example can be the p-waves in an earthquake.

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3 years ago

Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

trapecia [35]

Answer:

$CuCl_{2}(aq.)$ will form a precipitate of insoluble $CuCO_{3}$ when aqueous $Na_{2}CO_{3}$ is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and $NH_{4}^{+}$

all salts of sodium are soluble

When $Na_{2}CO_{3}$ is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of $Na_{2}CO_{3}$ into $CuCl_{2}(aq.)$ will result precipitation of insoluble $CuCO_{3}$

Reaction: $Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)$

4 0

3 years ago

In a chemical reaction, substrate molecule A is broken down to form one molecule of product B and one molecule of product C. The

MAVERICK [17]

Answer:

The answer to your question is d. 0.5 M

Explanation:

Data

[A] = 1M

K = 0.5

Concentration of B and C at equilibrium = x

Concentration of A at equilibrium = 1 - x

Equation of equilibrium

k = $\frac{[B][C]}{A}$

Substitution

$0.5 = \frac{[x][x]}{1 - x}$

Simplification

0.5 = $\frac{x^{2}}{1 - x}$

Solve for x

0.5(1 - x) = x²

0.5 - 0.5x = x²

x² + 0.5x - 0.5 = 0

Find the roots x₁ = 0.5 x₂ = -1

There are no negative concentrations so the concentration of A at equilibrium is

[A] = 1 - 0.5

= 0.5 M

4 0

3 years ago