The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:

Or,

where,
= 
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:

Hence, the freezing point of solution is 2.6°C
Answer:
domain bacteria
Explanation:
Salmonella and E. coli are same in the sense that they are both bacteria,
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.
Learn more about titration here:-brainly.com/question/186765
#SPJ4
Answer:
+ axit
CH2=CH-CH2-COOH,
CH3-CH=CH-COOH (tính cả đồng phân hình học)
CH2=C(CH3)-COOH.
+ este
HCOOCH=CH-CH3 (tính cả đồng phân hình học)
HCOO-CH2-CH=CH2,
HCOOC(CH3)=CH2.
CH3COOCH=CH2
CH2=CH-COOCH3