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kvv77 [185]
3 years ago
8

If the distance between two objects is decreased to - of the original

Chemistry
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

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The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
Why are the transition metals able to show different properties like varying color and paramagnetism ?
Naya [18.7K]

Answer:

Colors of transition metal compounds are due to two types of electronic transitions. Due to the presence of unpaired d electrons, transition metals can form paramagnetic compounds. Transition metals are conductors of electricity, possess high density and high melting and boiling points.

Explanation:

4 0
2 years ago
Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?
kifflom [539]
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction 
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl 
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0
3 years ago
Read 2 more answers
Through the complete electrolysis of a sample of pure water, a student collects 14.0 grams of hydrogen gas and 112.0 grams of ox
Vadim26 [7]

Answer:

126.0g of water were initially present

Explanation:

The electrolysis of water occurs as follows:

2H₂O(l) ⇄ 2H₂(g) + O₂(g)

<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>

<em />

To find the mass of water we need to determine moles of oxygen and hydrogen, thus:

<em>Moles Hydrogen:</em>

14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂

<em>Moles Oxygen:</em>

112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂

Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:

7 moles H₂O * (18g / mol) =

<h3>126.0g of water were initially present</h3>
6 0
3 years ago
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
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