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vodka [1.7K]
4 years ago
13

Which postulate of Dalton’s atomic theory has not been disproved by further investigation?

Chemistry
1 answer:
hram777 [196]4 years ago
5 0

Answer:

The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons...

hope this helps u

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The invisible force that surrounds a magnet is called the ________
Sladkaya [172]
Magnetic field
happy to help
8 0
3 years ago
Restaurant​ a's chicken salad contains the same number of grams of fat as the chicken salad of restaurant
german
To compute for grams of fat in each salad:

We know that, A = C.

And B is 5 grams more, so it is B = 5 + A

And the total fat is 65, so A + B + C = 65 grams.


Computation:

A + (5 + A) + A = 65

3A + 5 = 65

3A = 60

A = 20

A = 20

B = (65 - 40) = 25

C = 20
6 0
4 years ago
How many milliliters of a 1.25 molar hydrochloric acid solution would be needed to react completely with 60 grams of calcium
Fittoniya [83]
2HCl + Ca = CaCl₂ + H₂

n(HCl)=cv
n(Ca)=m(Ca)/M(Ca)
n(HCl)=2n(Ca)

cv=2m(Ca)/M(Ca)

v=2m(Ca)/{cM(Ca)}

v=2*60/{1.25*40}=2.4 L


7 0
3 years ago
Some of the following descriptions of a parrot are qualitative and some are quantitative. Which of the descriptions are qualitat
max2010maxim [7]

Answer:

1&3

Explanation:

because they are the ones without numbers :)

8 0
3 years ago
Read 2 more answers
At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction
Brums [2.3K]

Answer : The equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of  NO = 2 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NO = 2 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

                            2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Initial conc.          2 M            1 M            0

At eqm. conc.    (2-2x) M   (1-x) M         (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}

The K_c for reverse reaction = \frac{1}{1.6\times 10^{-5}}

Now put all the given values in this expression, we get :

\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of NO,Cl_2\text{ and }NOCl at equilibrium are :

Concentration of NO = (2-2x) M  = (2 - 2 × 0.975) M = 0.05 M

Concentration of Cl_2 = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of NOCl = x M = 0.975 M

Therefore, the equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

6 0
3 years ago
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