The concentration of the sodium hydroxide solution in mol/l is 0.176 M.
Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.
Given
V =25 ml = 0.025 L
M = 0.1
C₁ = 0.1
V₁ = 21.50 = 0.022 L
C₂ = ?
V₂ = 25 ml = 0.025 L
C₁V₁ = C₂V₂
C₂ = C₁V₁ / V₂
= 0.1 * 0.022 * 2 / 0.025
= 0.176 M
The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.
The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.
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First option.
The male tails are more attractive because of their long feathers , meanwhile the female tails are shorter than expected!
Answer:
Explanation:
The definition of acids and bases by Arrhenius Theory was modified and extended by Bronsted-Lowry.
Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.
In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.
From the given information:
From above:
We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid
accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.
The formula of the reactant that acts as a proton donor is <u>HCN</u>
The formula of the reactant that acts as a proton acceptor is <u>H2O</u>
Answer:
a. 50ml b.10ml c. 6.097ml d. 190.1 ml
Explanation:
According to Boyle's law
Volume is inversely proportional to pressure at constant temerature
Mathematically
P1V1=P2V2
P1=Initial pressure=0.8atm
V1=Initial volume=25ml
making V2 the subject
at 0.4atm P2=0.4 atm,
V2=25×0.8/0.4
=50ml
at 2 atm V2=25×0.8/2
=10 ml
1mmHg=0.00131579
2500mmHg=3.28 atm
At 3.28 atm,V2=25×0.8/3.28
=6.097 ml
at 80.0 torr
1 torr=0.00131579
80 torr=0.1052 atm
at 0.1048 atm V2=25×0.8/0.1048
=190.1 ml
Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.
