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3 years ago

What features of earth's crust do convergent divergent and transform boundaries form

Chemistry

1 answer:

12345 [234]3 years ago

8 0

Plate tectonics are the broken plates of earth that move around and form the different plate boundaries

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A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

= 21284.56606

mass of CaCO3 = 2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

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4 0

2 years ago

The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 + X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

8 0

3 years ago

Kendra conducts an experiment. She mixes two chemicals, resulting in a chemical reaction. Based on Kendra's

kondaur [170]

Answer: It changed in identity and properties.

Explanation:

3 0

3 years ago

Which of the following did your answer include?

IgorC [24]

Answer:

check all 4

Explanation:

3 0

3 years ago

Can someone please help me with this?--20 pts!

katrin2010 [14]

Answer:

Compound B is ionic. The electronegativity difference is 2.2, which can be determined by subtracting the electronegativity of Element Y from that of Element Z. Electronegativity differences greater than 1.7 indicate ionic bonds.

Hope that helps.

3 0

2 years ago