What are the building block of lipids?

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o

Sladkaya [172]

Potassium 23.5g/39.0983g/mol = 0.601mol

The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol

Therefore 0.3005mol of F2 is needed to find liters use

formula V = nRT/P (V)Volume = 22.41L

(T)Temperature = 273K or 0.0 Celsius

(P)Pressure = 1.0atm

(R)value is always .08206 with atm n = 0.3005moles (273)(.08206)(0.3005)/1 = V V = 6.7319 Liters

8 0

3 years ago

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Some properties of liquid water are very different from those of solid water (ice). Explain why freezing is considered a physica

drek231 [11]

Answer:

A chemical change is a change in the chemical makeup of a substance and results in a new substance. Ice and water both have the same chemical makeup - ice is just a different state and has the same atoms but in an expanded form.

Explanation:

8 0

3 years ago

Calculate the mass of ethyl alcohol required to prepare 540 grams of C4H6, if the reaction follows the scheme:

kkurt [141]

Answer:

Explanation:

2C₂H₅OH = C₄H₆ + 2H₂O + H₂

2 mole 1 mole

molecular weight of ethyl alcohol

mol weight of C₂H₅OH = 46 gm

mol weight of C₄H₆ 54 gm

540 gm of C₄H₆ = 10 mole

10 mole of C₄H₆ will require 20 mol of ethyl alcohol .

20 mole of ethyl alcohol = 20 x 46

= 920 gm

ethyl alcohol required = 920 gm .

3 0

3 years ago

What does combustion mean

frutty [35]

Answer:

combustion is a high-temperature exothermic redox chemical reaction between a fuel and an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke.

6 0

2 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$