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fredd [130]
3 years ago
12

Explain the environmental benefits of recyling paper products.

Chemistry
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

hi

Explanation:

bye

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if you help me ill give you more brainly's. plzz help.At 1 atm, how much energy is required to heat 37.0 g of H2O(s) at –10.0 °C
artcher [175]

Answer:

116.3 kJ

Step-by-step explanation:

Three heat transfers are involved

q = Heat to warm ice + heat to melt ice + heat to warm water + heat to evaporate water + heat to warm steam

q =      q₁      +     q₂        +     q₃       +     q₄          +     q₅

q = mC₁ΔT₁ + mΔH_fus + mC₃ΔT₃ + mΔH_vap + mC₅ΔT₅

<em>Step 1</em>: Calculate q₁

m = 37.0 g

C₁ = 2.010 J·°C⁻¹g⁻¹

ΔT₁ = T_f – T_i  

ΔT₁ = 0.0 – (-10.0)

ΔT₁ = 10.0 °C  

q₁ = 37.0 × 2.010 × 10.0  

q₁= 743.7 J

q₁= 0.7437 kJ

===============

<em>Step 2</em>. Calculate q₂

ΔH_fus = 334 J/g

q₂ = 37.0 × 334

q₂ = 12 360 J

q₂ = 12.36 kJ

===============

Step 3: Calculate q₃

C₃ = 4.179 J·°C⁻¹g⁻¹

ΔT₃ = T_f – T_i  

ΔT₃ = 100 – 0  

ΔT₃ = 100 °C

q₃ = 37.0 × 4.179 × 100  

q₃ = 15 460 J

q₃ = 15.46 kJ

===============

<em>Step 4</em>. Calculate q₄

ΔH_vap = 2260 J/g

q₄ = 37.0 × 2260

q₄ = 83 620 J

q₄ = 83.62 kJ

===============

<em>Step 5</em>. Calculate q₅

C¬₅ = 2.010 J·°C⁻¹g⁻¹

ΔT₅ = T_f – T_i  

ΔT₅ = 155.0 – 1000  

ΔT₅ = 55.0 °C

q₅ = 37.0 × 2.010 × 55

q₅ = 4090 J

q₅ = 4.090 kJ

===============

Step 6. Calculate q

q = 0.7437 + 12.36 + 15.46 + 83.62 + 4.090

q = 116.3 kJ

The heat required is 116.3 kJ.

4 0
3 years ago
Barbara is conducting an experiment to observe heat flow. She places one piece of metal in a freezer at -18 degrees Celsius (°C)
yulyashka [42]

<u>Answer:</u> The correct answer is heat will flow from the boiling water into the frozen metal.

<u>Explanation:</u>

According to the law of conservation of energy, energy cannot be destroyed nor created but it can be transformed from 1 form to another form.

There are 3 processes of heat transfer:

  1. <u>Conduction:</u> This type of heat transfer occurs when there is direct contact between the two objects.
  2. <u>Convection:</u> This type of heat transfer occurs when there is a movement of fluid (liquid or gas) due to the movement of hot layers to the top and cold layers to the bottom which leads to convection currents.  
  3. <u>Radiation:</u> This type of heat transfer occurs when there is a direct transfer of energy through space.

The heat moves from a hot surface to a cold surface to maintain equilibrium.

We are given:

A metal is placed in a freezer (cold object) and another metal is placed in an oven (hot object) and then both the metals are placed in boiling water.

Initially, the heat will flow from the boiling water (hot object) into the frozen metal (cold object) to maintain equilibrium

Hence, the correct answer is heat will flow from the boiling water into the frozen metal.

4 0
3 years ago
Calculate the density of an object with a mass of 3.8 g, that when placed in a 10.0 mL graduated cylinder with an initial volume
Arturiano [62]

Answer : The density of an object is 0.93 g/mL

Explanation : Given,

Mass of an object = 3.8 g

Initial volume = 4.5 mL

Final volume = 8.6 mL

First we have to calculate the volume of an object.

Volume of an object = Final volume - Initial volume

Volume of an object = 8.6 mL - 4.5 mL

Volume of an object = 4.1 mL

Now we have to calculate the density of an object.

Formula used:

\text{Density}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now putting all the given values in this formula, we get:

\text{Density}=\frac{3.8g}{4.1mL}

\text{Density}=0.93g/mL

Therefore, the density of an object is 0.93 g/mL

3 0
3 years ago
How many sigfigs in 10.0 ?<br> Your answer<br> I
alina1380 [7]

Answer:

3

Explanation:

10.0 has 3 significant figures

7 0
3 years ago
What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?
Katen [24]

Answer:

10.76 grams

Explanation:

Given that the amount of KMnO_4 is 25.0 grams.

Mass of 1 mole of KMnO_4 = 158 grams

The number of atoms in 1 mole of KMnO_4 is 4.

Mass of oxygen in 1 mole of KMnO_4 = 16\times 4 = 68 grams.

Here, 158 grams of KMnO_4 has 68 grams of oxygen

So, the amount of oxygen in 1 gram of KMnO_4 = 68/158 grams

Therefore,  the amount of oxygen in 1 gram of KMnO_4 = \frac {68}{158}\times 25 grams

=10.76 grams

Hence, 25 grams of KMnO_4 has 10.76 grams of oxygen.

3 0
2 years ago
Read 2 more answers
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