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Afina-wow [57]
3 years ago
14

Which statement best relates the strength and concentration of a base? At high enough concentrations, a weak base becomes strong

. Even at low concentrations, a strong base is strong. A weak base always has a low concentration. The stronger the base, the higher its concentration.
Chemistry
2 answers:
Lorico [155]3 years ago
6 0

Answer: Option (b) is the correct answer.

Explanation:

When there are more number of hydroxide ions in a solution then there will be high concentration of OH^{-} or hydroxide ions. As a result, more will be the strength of base in that particular solution.

A base is strong when it readily dissociate into its ions in the solution. When a base is strong, then it does not matter at what concentration it is dissolved in the solution because despite of its low concentration it will remain a strong base.

Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.


Anna35 [415]3 years ago
4 0
I think that right annswer for this question is statement number two.
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Balance this equation. if a coefficient of "1" is required, choose "blank" for that box.
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Step 1: Write Imbalance Equation

                            CH₃CHO  +  O₂    →    CO₂  +  H₂O

Step 2: Balance Carbon Atoms:
                                                 There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.

                             CH₃CHO  +  O₂    →    2 CO₂  +  H₂O 

Step 3: Balance Hydrogen Atoms:
                                                      There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.

                    CH₃CHO  +  O₂    →    2 CO₂  +  2 H₂O

Step 4: Balance Oxygen Atoms:
                                                   There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e  

                    CH₃CHO  +  5/2 O₂    →    2 CO₂  +  2 H₂O

Step 6: Eliminate Fraction:
                                         Multiply overall equation by 2 to eliminate fraction. i.e. 

                    2 CH₃CHO  +  5 O₂    →    4 CO₂  +  4 H₂O
7 0
3 years ago
PLEASE HELP ASAP. I WILL MARK BRAINLIEST!!!
iris [78.8K]

I Cant Answer your question but maybe this will help

Volume Changes for Gases

Particles in a gas have more freedom of movement than they do in a liquid. According to the ideal gas law, the pressure (P) and volume (V) of a gas are mutually dependent on temperature (T) and the number of moles of gas present (n). The ideal gas equation is PV = nRT, where R is a constant known as the ideal gas constant. In SI (metric) units, the value of this constant is 8.314 joules ÷ mole - degree K.

Pressure is constant: Rearranging this equation to isolate volume, you get: V = nRT ÷ P, and if you keep the pressure and number of moles constant, you have a direct relationship between volume and temperature: ∆V = nR∆T ÷ P, where ∆V is change in volume and ∆T is change in temperature. If you start from an initial temperature T0 and pressure V0 and want to know the volume at a new temperature T1 the equation becomes:

V1 = [n • R • (T1 - T0) ÷ P] +V0

Temperature is constant: If you keep the temperature constant and allow pressure to change, this equation gives you a direct relationship between volume and pressure:

V1 = [n • R • T ÷ (P1 - P0)] + V0

Notice that the volume is larger if T1 is larger than T0 but smaller if P1 is larger than P0.

Pressure and temperature both vary: When both temperature and pressure vary, the the equation becomes:

V1 = n • R • (T1 - T0) ÷ (P1 - P0) + V0

Plug in the values for initial and final temperature and pressure and the value for initial volume to find the new volume.

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Answer:

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