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1 year ago

Carbon (C): 1sH2sI2pJ H = I = J =

Chemistry

2 answers:

ratelena [41]1 year ago

4 0

Answer:H=2 I=2 J=2

Explanation:

Greeley [361]1 year ago

4 0

Answer:

H= 2

I= 2

J=2

Explanation:

Just did this answer on ingenuity. :)

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What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of

ololo11 [35]

Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

$Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995$

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

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1 year ago

By which process is a precipitate most easily separated from the liquid in which it is suspended

Gelneren [198K]

<span>Filtration, if its a precipitate that means its insoluble. </span>

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1 year ago

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Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alpha

klasskru [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is $E_{cell}__{AC}} = 0.94$

Explanation:

From the question we are told that

the cell voltage for AD is $E_{cell}__{AD}} = 1.56V$

From the data give we can see that

$E_{cell}__{AD}} - E_{cell}__{BD}} = E_{cell}__{AB}}$

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In the same way we can say that

$E_{cell}__{AD}}-E_{cell}__{CD}} = E_{cell}__{AC}}$

=> $E_{cell}__{AC}}=1.56- 0.62$

$E_{cell}__{AC}} = 0.94$

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1 year ago

If a chemical reaction such as the formation of iron oxide contains 4 atoms of iron (fe) in the product how many atoms of iron (

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4. Number of atoms are conserved in chemical reactions

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A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

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1 year ago