PLEASE HELP WITH THESE 3 QUESTIONS

Common element in all the organic compound another than hydrogen is

Leokris [45]

Answer:

Although all organic compounds contain carbon, and almost all have hydrogen, most of them contain other elements as well. The most common other elements in organic compounds are oxygen, nitrogen, sulfur, and the halogens.

Explanation:

I hope this helped!

5 0

4 years ago

Alex saw a rusty old car in an abandoned lot. Because he had recently read about chemical reactions, he began to wonder how the

Mice21 [21]

The answere is A .......

8 0

3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

4 years ago

Metric prefixes 9th grade level

zhannawk [14.2K]

Answer:

<h3>A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or submultiple of the unit. All metric prefixes used today are decadic. Each prefix has a unique symbol that is prepended to any unit symbol.</h3>

6 0

3 years ago

CO

KonstantinChe [14]

Answer:

1) The number of the listed compounds that have ionic bonds between them are 5 compounds

2) Their formulas showing their charges within the compounds are;

NaCl ↔ Na⁺+Cl⁻

MgO ↔ Mg²⁺+O²⁻

K₂O ↔ 2K⁺+O²⁻

LiF ↔Li⁺+F⁻

CaO ↔ Ca⁺+O²⁻

3) The major reason they were selected is because they are compounds formed by the combination of a metal and a nonmetal

Explanation:

1) An ionic bond involves the transfer of an atoms valence to another atom, such that the ions formed by the ionic chemical bond will be oppositely charged

Therefore, metals, which are elements that mainly have a few electrons in their outermost shell, loses their valence electrons and become positively charged, while non metals, which require a few electrons to complete the stable octet electronic configuration, receive the electrons from the metallic atoms and become negatively charged

Therefore, the compounds formed between a metal and a nonmetal in the list are the the compounds containing ionic bonds, including

NaCl, MgO, K₂O, LiF, and CaO

Therefore, 5 compounds contain ionic bonds between them;

2) Their formulas are;

NaCl, MgO, K₂O, LiF, and CaO

3) The major reason that the chosen compounds contain ionic bonding between them is because they involve the combination of a metal and a nonmetal as follows;

a) NaCl → Na⁺ (metal ion) + Cl⁻(nonmetal ion)

b) MgO → Mg²⁺ (metal ion) + O²⁻(nonmetal ion)

c) K₂O → 2K⁺ (metal ion) + O²⁻(nonmetal ion)

d) LiF → Li⁺ (metal ion) + F⁻(nonmetal ion)

e) CaO → Ca⁺ (metal ion) + O²⁻(nonmetal ion)

6 0

3 years ago