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2 years ago

What is waste good for?

Chemistry

1 answer:

harkovskaia [24]2 years ago

6 0

A because that honestly just makes the most sense

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An environmental scientist studies _______.

Assoli18 [71]

An environmental scientist studies the environment - you can see that in the beginning of the sentence :)

7 0

3 years ago

Read 2 more answers

If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?

rewona [7]

Answer:

When we finish, the temperature would be 32.5℃

Explanation:

Density of water = mass/volume

So,

Mass of water = Density × Volume

$\\\\$=1.0 \times 2.0 L$\\\\$=1.0 \frac{g}{m L} \times 2000 m L$\\\\$\quad=2000 g$$

$Q=m \times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

plugging in the values

$15000 \mathrm{Cal}=2000 \mathrm{g} \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times \Delta T$

$\\$\Delta T=\frac{15000 \mathrm{cal}}{2000 \mathrm{g} \times \frac{1.0 \mathrm{cal}}{g^{\circ} \mathrm{C}}}$\\\\$\Delta T=7.5^{\circ} \mathrm{C}$$

Final T = ∆T + Initial T

= 7.5℃ + 25℃ = 32.5℃ (Answer).

5 0

2 years ago

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

1 gallon = 3.785 L

1 L = 1000 mL

Now we convert 1.24 gallons to mL:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We calculate the mass of Octane:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

7 0

3 years ago

Have an infinite number of significant figures

vovikov84 [41]

The answer is: Non-zero

7 0

2 years ago

The maximum efficiency possible in an energy conversion process that is not limited by the second law of thermodynamics

Anna [14]

Answer:

100 %

Explanation:

The maximun efficiency possible (whem not limited by the second law of thermodynamics) happens when all the energy used is transformed into the type of energy we required with no other transformations.

For example, in an engine we want that all the energy we supply is being converted to work. That's the ideal case, but in reality always some of that energy is lost in the form of heat.

5 0

3 years ago