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2 years ago

How many moles of hydrogen are in 3.0 moles of C6H12O6?

Chemistry

2 answers:

tangare [24]2 years ago

8 0

<h3>Answer:</h3>

36 moles of Hydrogen

<h3>Solution:</h3>

The molecular formula of Glucose is,

C₆H₁₂O₆

As clear from molecular formula, each mole of Glucose contains 12 moles of Hydrogen atoms.

Therefore,

1 mole of C₆H₁₂O₆ contains = 12 moles of Hydrogen

So,

3.0 moles of C₆H₁₂O₆ will contain = X moles of Hydrogen

Solving for X,

X = (3.0 mol × 12 mol) ÷ 1 mol

X = 36 moles of Hydrogen

ki77a [65]2 years ago

3 0

Answer:

Explanation:

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

2 years ago

Ight i'm soo bored just wanna talk hit me up

ElenaW [278]

Answer:

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Explanation:

8 0

2 years ago

A solution is created by dissolving 13.5 grams of ammonium chloride in enough water to make 235 ml of solution. how many moles o

erica [24]

<span>Moles = 0.252 Molarity = 1.07 This question is badly worded. You're asking for moles and I suspect you really want molarity. The number of moles of ammonium chloride you have in the solution will remain constant regardless of the volume of the solution. However, the molarity of the solution will differ depending upon how concentrated it is. So I'll give you both the number of moles of ammonium chloride you have, and the molarity of the resulting solution. Please talk to your teacher if you're confused by the difference between moles and molarity. The formula for ammonium chloride is NH4Cl. So let's calculate it's molar mass. Start by looking up the associated atomic weights. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight chlorine = 35.453 Molar mass NH4Cl = 14.0067 + 4 * 1.00794 + 35.453 = 53.49146 g/mol Moles NH4Cl = 13.5 g / 53.49146 g/mol = 0.252376735 mol Molarity is defined as moles per liter, so let's divide the number of moles we have by the volume in liters. So: 0.252376735 mol / 0.235 l = 1.073943551 M Rounding to 3 significant figures gives: 0.252 moles, 1.07 molarity.</span>

7 0

3 years ago

An important step in science is supporting a theory or idea without data. The questions we ask determine the type of data we col

san4es73 [151]

Answer:

What will happen to Uk if you double the mass?

Explanation:

Uk = 0.5 * m * v²

You see that both m and v are variable, which means that both m and v can be any number. Regardless of the numbers you put in for m or v, the formula to calculate the kinetic energy (Uk) remains valid.

You could ask

1. What will happen to Uk if you double the mass?

2. What will happen to Uk if you double the velocity?

please see and understand(!) that the relationship between Uk an v² is indeed the velocity squared....

EXTRA

Uk = 0.5 * m * (v)²

Suppose the m = 3kg and velocity = 5 m/s

What is the Uk?

Well if you know the formula you can use your calculator to find out:

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (5)²

Uk = 0.5 * 3 * 25

Uk = 37.5 kgm/s²

Again you ask what will happen to Uk if you double the velocity?

At first it was 5 m/s and now it doubles, which means it now has that value *2

The new velocity is 5 *2 = 10 m/s

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (10)²

Uk = 0.5 * 3 * 100

Uk = 150 kgm/s²

150 = 4 * 37.5

So now you see that if you double your velocity, the Uk will be 2² = 4 times as big !

3 0

2 years ago

3. Which of the following are examples of chemical changes? Select all that apply.

Kruka [31]

Answer:

wood burning and cookies baking

Explanation:

took the test <333

3 0

2 years ago