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3 years ago

Br2 + 2LiF → 2LiBr + F2

Chemistry

1 answer:

N76 [4]3 years ago

6 0

1 grams F2 to mol = 0.02632 mol

10 grams F2 to mol = 0.26318 mol

20 grams F2 to mol = 0.52636 mol

30 grams F2 to mol = 0.78954 mol

40 grams F2 to mol = 1.05272 mol

50 grams F2 to mol = 1.3159 mol

100 grams F2 to mol = 2.6318 mol

200 grams F2 to mol = 5.2636 mol

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How many moles are in 3.45g of KCI ?

artcher [175]

<h3>Answer:</h3>

0.0463 mol KCl

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.45 g KCl

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

Molar Mass of K - 39.10 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.45 \ g \ KCl(\frac{1 \ mol \ KCl}{74.55 \ g \ KCl})$
Multiply/Divide: $\displaystyle 0.046278 \ mol \ KCl$

<u>Step 4: Check</u>

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

0.046278 mol KCl ≈ 0.0463 mol KCl

8 0

3 years ago

Write a balanced half-reaction for the reduction of gaseous nitrogen (N2) to aqueous hydrazine (N2H4) in basic aqueous solution.

weqwewe [10]

Answer:

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

Explanation:

The half reaction for the reaction for the reduction of gaseous nitrogen to aqueous hydrazine is;

N₂(g) → N₂H₄(aq)

The balancing the atoms in the half reaction. Hydrogen atom is balanced by adding hydrogen ions (H⁺)

We have;

N₂(g) + 4H⁺(aq) → N₂H₄(aq)

Then we balance the charge on both sides by adding electrons where the positive charge is greater;

we have;

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

5 0

4 years ago

Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv

poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

$Q_{1} = mC_{1} \Delta T_{1}$

Putting the given values into the above equation as follows.

$Q_{1} = mC_{1} \Delta T_{1}$

= $27.3 g \times 1.70 J/g K \times 41$

= 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

$Q_{2}$ = energy required = $mL_{v}$

$L_{v}$ = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

$Q_{2}$ = $mL_{v}$

= $27.3 \times 444$

= 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

$Q_{3} = mC_{2} \Delta T_{2}$

Value of $C_{2}$ = 1.06 J/g, $\Delta T_{2}$ = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

$Q_{3} = mC_{2} \Delta T_{2}$

= $27.3 g \times 1.06 J/g \times 34 K$

= 983.892 J

Therefore, net heat required will be calculated as follows.

Q = $Q_{1} + Q_{2} + Q_{3}$

= 1902.81 J + 12121.2 J + 983.892 J

= 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0

3 years ago

Compare the strength of intermolecular forces in no2 to the strength of intermolecular forces in n2

matrenka [14]

N2 is non-polar molecules,
NO2 is polar molecules that have slightly positive charge on N and slightly negative on O, so <span>intermolecular forces are stronger in NO2</span>

7 0

3 years ago

Read 2 more answers

How many moles are present in 88.0 g of KCN?

Kruka [31]

I think so x=1,35 moles

7 0

3 years ago