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maksim [4K]
3 years ago
9

How does the difference in the mass of the ball affects the changes in the distance of the object

Chemistry
1 answer:
beks73 [17]3 years ago
4 0

Answer:

In very simple words ,

As the weight increases the gravitational force acting on the ball increases and thus it effects the change in the distance of the ball as the ball will fall very close to its initial distance if it has more mass and shall go far if the mass is less .

You might be interested in
Why Mg ion +2 is smaller than Mg ion ​+1 ?
Bezzdna [24]

Answer:

Rise in charge reduced the size due to rise in attraction between electron and nucleus.

Explanation:

Ion of magnesium is smaller than the atom of magnesium. As the nuclear charge increases, the attraction of electron to the nucleus increases in charged ions form of any element. Due to this increase in charge of magnesium ion,  the size of ion reduces

No. of electrons in Mg = 12

No. of proton in Mg = 12

Where as

No. of electrons in Mg2+ = 12

No. of proton in Mg2+ = 10

6 0
3 years ago
It took 4.5 minutes for 1.0 l helium to effuse through a porous barrier. how long will it take for 1.0 l cl2 gas to effuse under
serious [3.7K]
When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9  & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
∴Cl2 will take 18.945 min for 1 L to effuse under identical conditions

4 0
4 years ago
Sort these elements into pairs that would most likely exhibit similar chemical properties Br ,Mg ,F ,Sr ,S ,O
Lapatulllka [165]
The elements in the periodic table are arranged in columns - elements in the same column exhibit similar properties.
Mg and Sr are in the same column (period I)
Br and F  (period XVII - 17)
S and O (period XVI - 16)
8 0
4 years ago
How are acids and ionic compounds similar?
11Alexandr11 [23.1K]

Answer: Ionic compounds are held together by the virtue of their opposing charges. Na+Cl- for example. If we consider Hg+(2Cl-)2, a mercuric chloride, the solubility is much less. Ba++(SO)4 Barium Sulphate, is highly insoluble; all differ by the relative attractiveness by Differing opposing charge(s).

Acids are very similar, consider Formic Acid, HCOOH, the simplest of the Carboxylic Acids. It dissociates more than say Benzoic Acid, C6H5-COOH. But neither disassociate as fully as Nitric Acid HNO3.

So the relative disassociation of the H+ (proton), or H3O+, (Hydronium ion), from any of these in water vary for a number of reasons we need not consider now.

Here is a “Tricky One!” (And very nasty). Take HF liquid or gas. This is one of the strongest acids on Earth - AS A LIQUID compound OR GAS. It will dissociate essentially near completion! Eat the floor, and is very dangerous.

NOW - HF (aqueous). The HF is in water. Very like HCl? NO! Why you may ask...The Electrophilic nature of Fluorine, “bathed in water, with an H+ all its own”, doesn’t let it go as easily!

HF is HIGHLY ordered in water, you can almost imagine a sort of “Hydrated matrix”, little HFs in endless rows...

BUT BE WARNED - even the aqueous HF is so reactive it will dissolve bone!

(I was told it was extremely painful; and did not appear to heal for weeks!)

Explanation: so, both types of compounds have a similarity, held together by the strength of their opposing charges or the degree of dissociation, (using water for simplicity).

That should do it.

8 0
3 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
4 years ago
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