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3 years ago

Predict which surface will melt ice faster. Wood or metal? Why?

Chemistry

2 answers:

Kay [80]3 years ago

8 0

Answer:

Explanation:

Some materials allow heat to pass through easily, while others do not. Even though the metal block feels colder than the wooden block, it transfers the heat it has to the ice cube more easily than the wooden block does, and this makes the ice cube melt faster.

zmey [24]3 years ago

8 0

Answer:

I believe wood wood melt it faster. Why, because of how compact it is. It would be warmer. Metal would be a cooler temperature.

Explanation:

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Which statement is true about crystal lattice energy

Arturiano [62]

Hey there!:

1) The additional stability that accompanies the formation of the networkCrystalline is measured as network enthalpy.
2) The reticular energy is the energy released when the solid Crystal isform from separate ions in the gaseous state. Always exothermic.
3) The enthalpy of the network depends directly on the size of the loads and conversely in the distance between the ions .

hope this helps!

3 0

3 years ago

Read 2 more answers

I want to know the steps.

Artyom0805 [142]

The answer for the following problem is described below.

Therefore the standard enthalpy of combustion is -2800 kJ

Explanation:

Given:

enthalpy of combustion of glucose(Δ $H_{f}$ of $C_{6}H_{12} O_{6}$ ) =-1275.0

enthalpy of combustion of oxygen(Δ $H_{f}$ of $O_{2}$ ) = zero

enthalpy of combustion of carbon dioxide(Δ $H_{f}$ of $CO_{2}$ ) = -393.5

enthalpy of combustion of water(Δ $H_{f}$ of $H_{2} O$ ) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ $H_{f}$ = ∈Δ $H_{f}$ (products) - ∈Δ $H_{f}$ (reactants)

$C_{6}H_{12} O_{6}$ (s) +6 $O_{2}$ (g) → 6 $CO_{2}$ (g)+ 6 $H_{2} O$ (l)

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ $H_{f}$ = 6 (-393.5) + 6(-285.8) - 0 + 1275

Δ $H_{f}$ = -2361 - 1714 - 0 + 1275

Δ $H_{f}$ =-2800 kJ

Therefore the standard enthalpy of combustion is -2800 kJ

7 0

3 years ago

If the mass is 6.5g, and the volume is 25mL.. what is the density? PLEASE ANSWER ASAP

julia-pushkina [17]

Answer:

p = 260 kilogram/cubic meter

Explanation:

ρ = $\frac{m}{V} \\$

= $\frac{6.5 g}{25mL}$

= 0.26 gram/milliliter

= 260 kilogram/cubic meter

7 0

3 years ago

Anhydrous CoCl2 is blue; CoCl2•6H2O is red. When CoCl2•6H2O is heated and water is clearly being driven off, it becomes uniforml

Sedaia [141]

Answer: On losing 6 moles of water, cobalt chloride forms unstable violet-coloured ions, before generating its stable blue-coloured anhydrous form.

Explanation:

The hydrated cobalt chloride loses its 6 water of crystallization, then dissociates into ions: cobalt ions and chlorine ions that appear violet, and quickly combined to form the stable anhydrous Cobalt chloride with blue colour.

4 0

3 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

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