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MA_775_DIABLO [31]
3 years ago
6

What effect does reducing your carbon footprint have on the environment?

Chemistry
1 answer:
marin [14]3 years ago
4 0

Answer:

reduces the greenhouse effect

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 How much energy is needed to raise the temperature of 125g of water from 25.0oC to 35.0oC?  The specific heat of water is 4.184
Anvisha [2.4K]

Hello!

To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.

In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.

Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.

35.0° C - 25.0° C = 10.0° C

We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.

q = 125 g × 4.184 J/g °C × 10.0° C

q = 523 J/°C × 10.0° C

q = 5230 J

Therefore, it will take 5230 joules (J) to raise the temperature of the water.

6 0
3 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
the mechanical advantage of a wheel and axle is the radius of the wheel divided by radius of the wheel divided by the radius of
denis-greek [22]
B 6 ft 2 in. = _____ ft
7 0
3 years ago
Read 2 more answers
What is true about all forces?
Feliz [49]

Answer:1) true

2) true

3) false

4)true

Explanation:

6 0
3 years ago
Read 2 more answers
How is an average mass different from a weighted mass
romanna [79]

In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.

Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.

<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>

<em>Weighted average (Method 1) </em>

Use the <em>numbers of each</em> gender (15 girls + 5 boys) ,

Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20

= 1120 kg/20 = <em>56 kg</em>.

If you put all the students on one giant balance, their total mass would be

1120 kg and the average mass of a student would be <em>56 kg. </em>

<em>Weighted average (Method 2) </em>

Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).

Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>

Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.

6 0
3 years ago
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