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Stella [2.4K]
3 years ago
12

Please select the word from the list that best fits the definition

Chemistry
1 answer:
den301095 [7]3 years ago
8 0

Answer:

carboxylic acid

zhhzhdhdjhd

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Dolomite, [CaMg(CO3)2] is found in a soil sample. A geochemist titrates 24.65 g of soil with 57.85 mL of 0.3315M HCl. What is th
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In order to find out the %mass dolomite in the soil, calculate for the mass of dolomite using the information given from the titration procedure. You would need to multiply 57.85 ml with 0.3315 M HCl and you would get the amount of HCl in millimoles. Then multiply the amount of HCl with 1/2 (given that for every 1 mol of dolomite, 2 mol of HCl would be needed). Convert the amount of dolomite to mass by multiplying the millimoles with the molecular weight which is 184.399. Then convert the mass to grams which is 1.768 grams. Divide the mass of dolomite (1.768 grams) with the weight of soil sample. The % mass is 7.17. 

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3 years ago
What is an individual living entity called?
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1 year ago
Which of the following natural resources cant be reused
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What is the concentration (m) of sodium ions in 4.57 l of<br> a.847 m na3p solution?
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A group of environmentalists were discussing the benefits and drawbacks associated with using fossil fuels. Which argument <span>best </span>fits the conversation? 
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8 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
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