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2 years ago

What types of compounds are CaCl2, Cu, C2H6, respectively.

Chemistry

1 answer:

mina [271]2 years ago

5 0

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 ionic compound.

-) $Cu$

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 metal.

-) $C_2H_6$

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 organic compounds.  (See figure 1)

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2 years ago

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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2 years ago

Which of the following statements about monosaccharide structure is true?

Nana76 [90]

Answer:

The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)

Explanation:

Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5

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2 years ago

A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of sulfur, and 64.00 g of oxygen. How many grams of sulfur and grams of o

Gemiola [76]

From the chemical formula of sulfuric acid, we can see the molar ratio:

H : S : O
2 : 1 : 4

Now, we convert the mass of hydrogen given into the moles of hydrogen. This is done using

Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27

Therefore, the moles will be:

S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles

Now, the respective masses are:
S = 32 * 3.64 = 116.48 grams
O = 16 * 14.54 = 232.64 grams

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2 years ago

The diagram shows the movement of particles from one end of the container to the opposite end of the container. A cylindrical co

trapecia [35]

Answer:

C. effusion because there is a movement of a gas through a small opening into a larger volume

Explanation:

Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.

This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.

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