UNLIKE kinetic energy, potential energy is *

What is the freezing point of a solution that contains 36.0 g of glucose ( ) in 500.0 g of water? ( for water is 1.86°c/m. the m

Marizza181 [45]

Answer: - 1.86°C

Explanation:

The depression of freezing points of solutions is a colligative property.

That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.

To solve the problem follow these steps:

Data:

Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol

2) Formulas

Tf = Normal Tf - ΔTf

ΔTf = i * kf * m

m = number of moles of solute / kg of solvent

number of moles of solute = mass in grams / molar mass

3) Solution

number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol

m = 0.2 mol / 0.5 kg = 1.0 m

ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C

Tf = 0°C - 1.86°C = - 1.86°C

Answer: - 1.86 °C

7 0

3 years ago

Which of the following forms the greatest number of strong covalent bonds and is therefore the backbone of organic molecules inc

solniwko [45]

The correct answer for the given question above would be option B. HYDROGEN. The form that has the greatest number of strong covalent bonds and is therefore the backbone of organic molecules including carbohydrates, proteins, lipids and nucleic acids is HYDROGEN. Hope this answers your question.

8 0

3 years ago

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:

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

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

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

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

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

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

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

8 0

3 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 4.00 cm radius?

SOVA2 [1]

Answer:ahaha idk

Explanation:

6 0

3 years ago

What Are Forces?

Black_prince [1.1K]

Answer:

forces is a push or pull upon an object which the object is being interacted

dont see the following objects but made a black hole or a huge object or a hevey object

Explanation:

6 0

3 years ago

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