It's 1. Melting a substance. The rest are chemical changes
C = 4 mol/l
v = 0.5 l
n(NaCl)=cv
n(NaCl) = 4 mol/l · 0.5 l = 2 mol
2 moles of NaCl must be dissolved
The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is <u>3.347</u>.
Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.
<u>Calculation:-</u>
Normality of acid Normality of base
= nMV nMV
= 1 × 0. 15 × 0.017 1 × 0. 20 ×0.015 L
= 2.55 × 10⁻³ = 3 × 10⁻³
The overall base will be high
net concentration = 3× 10⁻³ - 2.55 × 10⁻³
= 0.45 × 10⁻³
= 4.5× 10⁻⁴
pH = -log[4.5 × 10⁻⁴]
= 4 - log4.4
= <u>3.347</u>
A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.
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Answer:
You are not using properly the function exponential in your calculator
Explanation:
When a number is too big or too small we use scientific notation. This is a number between 1 a 10 multiplied by a power of 10.
When you are writing 4.5*10^-9 you are actually writing 0.0000000045 in scientific notation.
When you enter this in the calculator you have to use the function EXP after the first two numbers.
Steps: 1) Enter 4.5
2) Enter EXP
3) Enter minus (-)
4) Enter 9
Answer:
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements. All elements are pure substances.
Explanation:
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