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Artemon [7]
2 years ago
15

There are three sets of sketches below, showing the same pure molecular compound (hydrogen chloride, molecular formula ) at thre

e different temperatures. The sketches are drawn as if a sample of hydrogen chloride were under a microscope so powerful that individual atoms could be seen. Only one sketch in each set is correct. Use the slider to choose the correct sketch in each set. You may need the following information: melting point of : -114.8 boiling point of : -85.1

Chemistry
1 answer:
RideAnS [48]2 years ago
8 0

Answer:

The correct answer is - option C.

Explanation:

Given: the melting point of HCl is

-114.8 °C, which suggests that below this temperature HCl will be solid.

and, since the boiling point of HCl is - 85.1 °C. It is also suggested that above this temperature HCl will be gas, Therefore.

Solid -114.8  - Ordered arrangement

Liquid -85.1c  - Less orderly arranged

Gaseous - Least orderly arranged

Thus, at —90 °C, HCl will be present 'in the liquid state,  At — 1 °C, HCl will be present in the gaseous state and at -129 °C, HCl will be present in the solid-state. So, the molecules will be organized in a more orderly manner .

Thus, the correct answer is - option C

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If you weigh 100 kg, how much would you weigh if all the water were removed from your body? A65 kg B45 kg C50 kg D35 kg
Mekhanik [1.2K]

Answer:

B. 45k

The human body is about 60 to 70% water.

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5 0
2 years ago
What types of compounds are CaCl2, Cu, C2H6, respectively.
mina [271]

Answer:

Ionic, metal, organic

Explanation:

In this case, we have to analyze each compound:

-) CaCl_2

In this compound, we have a non-metal atom (Cl) and a metal atom (Ca) . So, we will have a high electronegativity difference between these atoms, With this in mind, we will have an ionic bond. Ions can be produced:

CaCl_2~->~Ca^+^2~+~2Cl^-

The cation would be Ca^+^2 and the anion is Cl^-. So, we will have an <u>ionic compound.</u>

-) Cu

In this case, we have a single atom. If we check the periodic table we will find this atom in the transition metals section (in the middle of the periodic table). So, this indicates that Cu (Copper) is a <u>metal.</u>

-) C_2H_6

In this molecule, we have single bonds between carbon and hydrogen. The electronegativity difference between C and H are not high enough to produce ions. So, with this in mind, we will have covalent bonds. This is the main characteristic of <u>organic compounds. </u> (See figure 1)

5 0
2 years ago
Simplify: <br>(100 m)/(26 s)
podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

4 0
2 years ago
Read 2 more answers
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0
2 years ago
Read 2 more answers
A precipitate of zinc hydroxide can be formed using the reaction below.
worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH  = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0
2 years ago
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