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Deffense [45]
3 years ago
10

Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is requi

red to completely neutralize a 4.3 billion liter lake with a ph of 5.5
Chemistry
1 answer:
svet-max [94.6K]3 years ago
4 0
From the given pH, we calculate the concentration of H+:
     [H+] = 10^-pH = 10^-5.5
We then use the volume to solve for the number of moles of H+:
     moles H+ = 10^-5.5M * 4.3x10^9 L =  13598 moles
From the balanced equation of the neutralization of hydrogen ion by limestone written as
     CaCO3(s) + 2H+(aq)  → Ca2+(aq) + H2CO3(aq)
we use the mole ratio of limestone CaCO3 and H+ from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H+, to compute for the mass of the limestone:
     mass CaCO3 = 13598mol H+(1mol CaCO3/2mol H+)
                               (100.0869g CaCO3/1mol CaCO3)(1kg/1000g) 
                            = 680 kg
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How many moles of tungsten atoms are in 4.8x10^25 atoms of tungsten?
allsm [11]
Number of moles = number of atoms/Avogadro's number
= (4.8x10^25)/(6.02x10^23)
=79.7 mol (3sf)
7 0
3 years ago
¿En qué grupo de sustancias químicas se incluye a la Hemoglobina?
olga2289 [7]

Answer:

"HEMOGLOBINA. Las características de la hemoglobina (Hb) como amortiguador están íntimamente vinculadas a la capacidad de disociación del grupo imidazólico del aminoácido histidina unido al hierro que contiene el grupo hemo (ver figura inferior)."

Link: http://www.ehu.eus/biomoleculas/buffers/buffer4.htm#:~:text=HEMOGLOBINA&text=Las%20caracter%C3%ADsticas%20de%20la%20hemoglobina,hemo%20(ver%20figura%20inferior).

3 0
3 years ago
Calculate the volume a solution required toof provide the following: (a) 2.14 g of sodium chloridefrom a 0.270 M solution, (b) 4
enyata [817]
A) NaCl = 58.44 g/mol

number of moles :  2.14 / 58.44 = 0.0366 moles

V = n / M  => 0.0366 / 0.270 = 0.1355 L 
__________________________________________

b) C2H6<span>O = 46.06 g/mol

</span>number of moles :  4.30 / 46.06 = 0.0933 moles

<span>V = n / M  => 0.0933 / 1.50 = 0.0622 L 
</span>___________________________________________

c) CH3COOH = 60.0 g/mol

number of moles :  0.85 / 60.0= 0.0141 moles

V = n / M  =>  <span> 0.0141 </span>/ 0.30 = 0.047 L 
___________________________________________

hope this helps!
3 0
3 years ago
Barium nitrate (Ba(NO3)2) reacts with sodium chloride (NaCl) in a double replacement (displacement) reaction, shown below.
Nesterboy [21]

Answer:

you know that they will be a displacement reaction that will form a barium salt:

Ba(NO3)2+ 2NaCl--> BaCl2 + 2NaNO3

So now that we have that formula and the molecular weight we can determine how much salt will be made. So here we convert the grams to moles

(42.3g Ba(NO3)2)*(1 mole/261.34g) = 0.16185 mol

In the molecular formula we know that 1 mole of Barium nitrate will create 1 mole of Barium chloride, so in this case (in a perfect world) you should get 0.16185 mole of barium chloride (208.23 g/mol) that we then have to convert to grams.

(0.16185 mol BaCl2) * ( 208.23 g/mol) = 33.7037 g of Barium Chloride (rounded to 3 significant digits = 33.7g)

3 0
1 year ago
At a certain temperature this reaction follows second-order kinetics with a rate constant of Suppose a vessel contains at a conc
Norma-Jean [14]

The question is incomplete, here is the complete question:

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹

2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.

<u>Answer:</u> The concentration of SO_3 in the vessel after 0.240 seconds is 0.24 M

<u>Explanation:</u>

For the given chemical equation:

2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 14.1M^{-1}s^{-1}

t = time taken= 0.240 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 1.44 M

Putting values in above equation, we get:

14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)

[A]=0.245M

Hence, the concentration of SO_3 in the vessel after 0.240 seconds is 0.24 M

4 0
3 years ago
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