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Oksanka [162]
3 years ago
7

Newton contributed to astronomy by:

Chemistry
2 answers:
SCORPION-xisa [38]3 years ago
6 0

Answer:  D

Explanation:

This is the answer because everyone knows he discovered gravity and he conducted scientific experiments to prove them which he also used math for

Hope this helps

stiv31 [10]3 years ago
6 0

Answer:

Newton contributed to astronomy by <u><em>developing theories of gravity and motion.</em></u>

Explanation:

Newton reflected on the fact that bodies weighed on Earth and that the stars revolved around other stars. He imagined that there was a universal force that caused bodies to attract each other. Newton called it "Universal Gravitation Force" or "Gravity."

Then, the force of gravity is the physical force exerted by the mass of a planet on the objects that are within its gravitational field. In other words, this law establishes that bodies, by simply having mass, experience a force of attraction to other bodies with mass.

Objects are attracted according to their mass and the distance between their centers: if the distance increases, the force decreases, and if the mass increases, the Universal Gravitation Force is more intense:

F=G*\frac{M*m}{r^{2} }

where G is the universal gravitation constant, whose value is

This expression indicates that the gravitational force that a point mass M exerts on another point mass m that is at a distance r from the previous one is directly proportional to the product of the masses and inversely proportional to the square of the distances that separates them.

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3Al + 3 NH4ClO4 ---&gt; Al2O3 + AlCl3 + 3 NO + 6H20
Tresset [83]

Answer:

9.63 L of NO

Explanation:

We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:

Mass of NH₄ClO₄ = 50 g

Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)

= 14 + 4 + 35.5 + 64

= 117.5 g/mol

Mole of NH₄ClO₄ =?

Mole = mass /molar mass

Mole of NH₄ClO₄ = 50/117.5

Mole of NH₄ClO₄ = 0.43 mole

Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:

3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O

From the balanced equation above,

3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.

Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.

Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:

1 mole of NO = 22.4 L

Therefore,

0.43 mole of NO = 0.43 × 22.4

0.43 mole of NO = 9.63 L

Thus, 9.63 L of NO were obtained from the reaction.

6 0
3 years ago
In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat
Vaselesa [24]

Answer:

OH−(aq), and H+(aq)

Explanation:

Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;

MnO4^-(aq) + 8H^+(aq) +5e-------> Mn^2+(aq) + 4H2O(l)

The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.

For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.

Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.

4 0
3 years ago
The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.
user100 [1]

Answer:

c    43.38 g

Explanation:

The reaction  between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.  

For the fact that HCl  gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess,  while the limiting reagent should be MnO2 .  

Thus, the theoretical yield of Cl2 will be  60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),  

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% *  60.25 = 43.38 g

5 0
3 years ago
How much time would it take for 5.2 x 10^5 atoms of fermium-253 (half-life 3 days) to decay to 6.5 x 104 atoms?
Jlenok [28]

Answer:

9 days, or 3 half-lives

Explanation:

5.2x10^5=520000

6.5x10^4=65000

65000/520000=1/8, or 3 half-lives

3x3=9 days

I'm not the greatest at Chem but this seems more like math than Chem :)

4 0
3 years ago
A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber
In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g                 (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532          (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3   = (mass of BaCO3 )/(mass of sample )*100

                                                         = 0.242/(0.846 )*100

                                                         = 28.605%

5 0
4 years ago
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