CHEMISTRY COLLEGE

Write the name of the metric system prefix associated with each mathematical factor. Do not use any hyphens in your answers and do not abbreviate the prefixes. Spelling counts. 10^-6 ______________ 1/100 ______________ 0.001 _____________ 1000 ______________

Answers

Answer 1
Answer:

Answer:

1. micro

2. centi

3. milli

4. kilo

Explanation:

Prefixes, in mathematical exponents, are defined as the alphabetical names associated with the mathematical factors. Some examples are as follows:

10⁶ = Mega

10⁹ = Giga

10⁻⁹ = Nano

10⁻¹² = Pico

So, in order to name the metric system prefixes associated with each of the given mathematical factor, we first convert the factor into exponent form and then write its prefix, as follows:

10⁻⁶ = micro

1/100 = 0.01 = 1 x 10⁻² = centi

0.001 = 1 x 10⁻³ = milli

1000 = 1 x 10³ = kilo

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What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m.

Answers

Answer:

freezing point   (°C) of the solution =  - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3  

K_f = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

molality = (mole \ of \ solute )/(kg \ of \ solvent )

which can now be re-written as :

molality = (mole \ of \ Ca(NO_3)_2)/(kg \ of \ water)

molality = ((mass \ of \ \ Ca(NO_3)_2)/(molar \ mass of \ Ca(NO_3)_2) )/(kg \ of \ water)

molality = ((11.3 \ g )/(164 \ g/mol) )/(0.115 \ kg )

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

\Delta T =3 * (1.86 \ ^0C/m) * (0.599 \ m)

\Delta T =3.34^0 \ C

\Delta T = the freezing point of water - freezing point of the solution

3.34° C = 0° C -  freezing point of the solution

freezing point  (°C) of the solution =  0° C - 3.34° C

freezing point   (°C) of the solution =  - 3.34° C

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Answers

Answer:

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How much energy is released when 33.8 g of water freezes? The heat of fusion for water is 6.02 kJ/mol.

Answers

Answer:

11.29Kj

Explanation:

1. find moles of 33.8g of water

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Answers

Answer:

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