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2 years ago

A reliable source of information on groundwater is always

Chemistry

2 answers:

never [62]2 years ago

7 0

Up to date.

Hope this helps

balandron [24]2 years ago

3 0

Http://water.ky.gov/groundwater/Pages/GroundwaterAwareness.aspx

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never [62]

Go charge ur phone!!

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2 years ago

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Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.

Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0

2 years ago

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What is the molar mass of sulfur (S)?

Lesechka [4]

Answer:

A. 32.06 g/mol

Explanation:

The molar mass units are always g/mol

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2 years ago

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Can someone answer 4 for me please?

Mice21 [21]

I can’t see the whole question......

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3 years ago

How many moles are in 1.42 x 10^25<br> molecules of NaCl?

Sholpan [36]

<h3>Answer:</h3>

23.6 mol NaCl

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.42 × 10²⁵ molecules NaCl

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.42 \cdot 10^{25} \ molecules \ NaCl(\frac{1 \ mol \ NaCl}{6.022 \cdot 10^{23} \ molecules \ NaCl})$
Multiply/Divide: $\displaystyle 23.5802 \ mol \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

23.5802 mol NaCl ≈ 23.6 mol NaCl

3 0

2 years ago