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3 years ago

U

Chemistry

1 answer:

guapka [62]3 years ago

4 0

Answer:

fluoride ion with a charge of -1

Explanation:

If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

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How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?

lesya692 [45]

Answer:

3.72 mol Hg

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Density = Mass over Volume

Explanation:

Step 1: Define

D = 13.6 g/mL

54.8 mL Hg

Step 2: Identify Conversions

Molar Mass of Hg - 200.59 g/mol

Step 3: Find

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

Step 4: Convert

$745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} )$ = 3.71544 mol Hg

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.71544 mol Hg ≈ 3.72 mol Hg

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3 years ago

How many independent variables are allowed in an experiment

Dmitriy789 [7]

one independent variable

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2 years ago

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Which of the following is a polyatomic ion? O A. HCI O B. Na O C. PO, O D. 52

timurjin [86]

I'm not the best with science but I believe the answer is B. Na.

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3 years ago

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Somebody knows a good explanation for quantic numbers in chemistry???

taurus [48]

Answer:

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Explanation:

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3 years ago

When 60 mL of 0.22 M NH4Cl is added to 60 mL of 0.22 M NH3, relative to the pH of the 0.10 M NH3 solution the pH of the resultin

densk [106]

Answer:

Will be more acidic

Explanation:

The equilibrium of NH3 in water is:

NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).

Where equilibrium constant, Kb, is:

Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]

From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:

1.85x10⁻⁵ = [X] [X] / [0.10M]

1.8x10⁻⁶ = X²

X = 1.34x10⁻³ = [OH⁻]

As pOH = -log[OH⁻] = 2.87

And as pH = 14 - pOH

pH of the 0.10M NH3 is 11.13

Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:

pOH = pKb + log [NH4⁺] / [NH3]

Where pKb is -log Kb = 4.74 and [] are moles of both compounds.

Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:

pOH = 4.74 + log [0.0132] / [0.0132]

pOH = 4.74

pH = 14 - 4.74 = 9.26

That means the pH of the resulting solution will be more acidic

7 0

3 years ago