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3 years ago

Which is an example of an external cost

2 answers:

8 0

Hello there,
Air pollution from burning fossil fuels.
On a side note: External cost is also known as a negative externality

Hope this helps :))

~Top

mestny [16]3 years ago

5 0

Noise pollution and does not compensate

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For the neutralization reaction involving HNO3 and Ca(OH)2, how many liters of 1.55 M HNO3 are needed to react with 45.8 mL of a

liraira [26]

2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
mols Ca(OH)2 = M x L = ? 
Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols HNO3. 
Then M HNO3 = mols HNO3/LHNO3. You have mols and M, solve for L.

3 0

3 years ago

Read 2 more answers

Soo Hwan found a flowering plant at school and used a classification key to help him classify the plant. The flowers of the plan

KIM [24]

Answer:

Black eyed susan

Explanation:

...

8 0

3 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

algol13

Answer:

1) C₃H₄N

Explanation:

The empirical formula of a compound is the formula that gives the positive integer ratio of the atoms of the elements in the compound in the simplest form

The mass of carbon in the compound = 90 grams

The molar mass of carbon = 12.011 g/mol

The number of moles of carbon = 90 g/(12.011 g/mol) ≈ 7.4931313 moles

The mass of hydrogen in the compound = 11 grams

The molar mass of hydrogen = 1.00794 g/mol

The number of moles of carbon = 11 g/(1.00794 g/mol) ≈ 10.913348 moles

The mass of nitrogen in the compound = 35 grams

The molar mass of nitrogen = 14.0067 g/mol

The number of moles of carbon = 35 g/(14.0067 g/mol) ≈ 2.49880414 moles

Dividing by the smallest mole ratio gives;

The proportion of carbon, C = 7.4931313/2.49880414 = 2.9987 ≈ 3

The proportion of nitrogen, N = 10.913348 /2.49880414 = 4.367 ≈ 4

The proportion of nitrogen, N = 2.49880414 /2.49880414 = 1

Therefore, the empirical formula of the compound is C₃H₄N.

7 0

3 years ago

2Ag(s) + S(s) -> Ag2S(s)

ArbitrLikvidat [17]

Answer:

Sulphur

Explanation:

Sulphur because we been 2 ag for 1 s therefore 0.312 S will need 0.624 of Ag

6 0

2 years ago

Copper exists naturally as two isotopes copper-63 with a mass of 62.93 amu and copper-65 with a mass of 64.93 amu. The average m

Andrej [43]

Answer: The percentage abundance of $_{29}^{63}\textrm{Cu}$ isotope is 69 %.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

Let the fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope be 'x'

For $_{29}^{63}\textrm{Cu}$ isotope:

Mass of $_{29}^{63}\textrm{Cu}$ isotope = 62.93 amu

Fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope = x

For $_{29}^{65}\textrm{Cu}$ isotope:

Mass of $_{29}^{65}\textrm{Cu}$ isotope = 64.93 amu

Fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope = (1-x)

Average atomic mass of copper = 63.55 amu

Putting values in above equation, we get:

$63.55=[(62.93\times x)+(64.93\times (1-x))]\\\\x=0.69$

Converting this fractional abundance into percentage abundance by multiplying it by 100, we get:

$\Rightarrow 0.69\times 100=69\%$

Hence, the percentage abundance of $_{29}^{63}\textrm{Cu}$ isotope is 69 %.

8 0

4 years ago