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Karolina
11 days ago
6

Two different ionic compounds each contain only copper and chlorine. Both compounds are powders, one white and one brown. An ele

mental analysis is performed on each powder. Which of the following questions about the compounds is most likely to be answered by the results of the analysis?
A. What is the density of each pure compound?
B. What is the formula unit of each compound?
C. What is the chemical reactivity of each compound?
D. Which of the two compounds is more soluble in water?
Chemistry
1 answer:
alisha [964]11 days ago
5 0

Answer:

B.

Explanation:

Elemental analysis reveals the precise percentage make-up of each component in the compound sample. Here, we have two distinct compounds, both containing copper and oxygen. However, the specific amounts of copper and oxygen in these compounds remain unknown.

An elemental composition will clarify this aspect and aid in determining the chemical formula for each compound, since the percentage of each element is now understood.

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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
KiRa [982]

Answer:

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

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles 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

Assuming mixing doesn't alter the total volume.

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 is calculated as moles of solute per kilograms of solvent.

Total 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 $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
1 day ago
You are given a piece of paper and a match. The paper has a mass of 2.5 g. You then light the match and light the piece of paper
KiRa [982]

Answer:

No

Explanation:

No. The demonstration in question does not infringe upon the conservation of mass.

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction; however, mass can change from one form to another during the process.

In this instance, although the remnants of the paper weigh 0.5 g compared to the original weight of 2.5 g, the ashes and gases produced during combustion account for the missing mass of the paper.

The portion that has been burnt has transformed into other states. If the gas and ashes are adequately contained, they will correspond to the weight of the original paper when added to the remaining paper.

5 0
1 day ago
Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi
Anarel [852]

The mixture’s density is 1.57 g/cm³.


Step 1: Determine the mass of the butter.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2: Determine the mass of the sand.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

6 0
1 day ago
The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um
lions [1003]

\boxed{\sf 1Å=10^{-10}m}

\\ \rm\longmapsto 2.27Å

\\ \rm\longmapsto 2.27\times 10^{-10}m

\\ \rm\longmapsto 0.227\times 10^{-9}m

\\ \rm\longmapsto 0.0227\times 10^{-8}m

\\ \rm\longmapsto 0.00227\times 10^{-7}m

\\ \rm\longmapsto 0.00023\times 10^{-6}m

\\ \rm\longmapsto 0.00023\mu m

6 0
12 days ago
1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?
alisha [964]

Answer: The correct selection is (b).

Explanation:

The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.

This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.

Therefore, a significant amount of energy is needed to dislodge the valence electrons.

The electronic configuration for helium is 1s^{2}. Hence, due to its fully occupied valence shell, it exhibits greater stability.

Consequently, a large amount of energy is needed to remove an electron from a helium atom.

In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.

7 0
9 days ago
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