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3 years ago

Use the following equation to answer question 1-4. Make sure you balance first.

Chemistry

1 answer:

worty [1.4K]3 years ago

4 0

<h3>Answer:</h3>

5.2 mol H₂O

<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>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

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

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})$
Multiply/Divide: $\displaystyle 5.2 \ mol \ H_2O$

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What is the heat of reaction (AH) of the<br> following?<br> C(s) + O2 (g) --> CO2 (g)

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Answer:

C co2 2co enthalpy

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4 years ago

The average speed of oxygen molecules in air is about ____. (1 point) 0 km/h 170 km/h 1700 km/h 17,000 km/h

LenKa [72]

The average speed is calculated using the following formula $speed = \sqrt{\frac{3kT}{m} }$

Where

$k = 1.381 X 10^{-23} J / K$

T = 298 K [room temperature]

m = mass of each molecule of oxygen

mass of one mole of oxygen molecules = 32 g / mol

mass of one molecule of oxygen will be = $\frac{32}{6.023 X 10^{23} }$

mass of one molecule of oxygen will be = $5.31 X 10^{-23} g = 5.31 X 10^{-26} kg$

putting values

average speed = $\sqrt{\frac{3 X 1.381 X 10^{-23} X 298}{5.31 X 10^{-26} kg} }$

Average speed = 482.19 m / s

average speed = 1735.8 km / h

so approx = 1700 km/h

4 0

3 years ago

Read 2 more answers

What is the temperature in degrees Celsius on the thermometer below?

ivolga24 [154]

Answer:

$\pmb{ option ~C}$

Explanation:

The correct answer of this questions is 38°c

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3 years ago

Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan

Valentin [98]

Answer : The equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

Explanation :

The given main chemical reaction is:

$N_2(g)+2O_2(g)\rightarrow N_2O_4(g)$ ; $K$

The intermediate reactions are:

(1) $N_2O_4(g)\rightarrow 2NO_2(g)$ ; $K_a$

(2) $\frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g)$ ; $K_b$

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) $2NO_2(g)\rightarrow N_2O_4(g)$ ; $\frac{1}{K_a}$

(2) $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$ ; $(K_b)^2$

Thus, the equilibrium constant for this reaction will be:

$K=\frac{1}{K_a}\times (K_b)^2$

$K=\frac{(K_b)^2}{K_a}$

Thus, the equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

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3 years ago

How many atoms are in 25.00 g of B.

iren2701 [21]

Answer:

$\boxed {\boxed {\sf 1.393 *10^{24} \ atoms \ B}}$

Explanation:

<u>1. Convert Grams to Moles</u>

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

Boron (B): 10.81 g/mol

Use this value as a ratio.

$\frac {10.81 \ g \ B }{1 \ mol \ B}$

Multiply by the given number of grams.

$25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}$

Flip the ratio so the grams of boron cancel out.

$25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}$

$25.00 *\frac {1 \ mol \ B }{10.81 }$

$\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B$

<u>2. Convert Moles to Atoms</u>

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

$\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

Multiply by the number of moles we calculated.

$2.312673451 \ mol \ B *\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

The moles of boron cancel.

$2.312673451 *\frac {6.02*10^{23} \ atoms \ B} {1 }$

$2.312673451 *6.02*10^{23} \ atoms \ B} =1.39269195*10^{24} \ atoms \ B$

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

$1.392\underline69195*10^{24} \ atoms \ B$

The 6 tells us to round the 2 to a 3.

$1.393 *10^{24} \ atoms \ B$

25.00 grams of boron is equal to 1.393*10²⁴ atoms.

6 0

3 years ago