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vovangra [49]
2 years ago
10

Green algae are often preserved as a fossil because they:

Chemistry
1 answer:
natita [175]2 years ago
5 0

Explanation:

I think

D. secrete slime

hope this helps you

have a nice day :)

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Do parts a, b and c
Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

pH=Pka+log(\frac{base}{acid})

pKa can be calculated from given Ka value as:

pKa=-logKa

pKa=-log(6.3*10^-^5)

pKa = 4.20

let's plug in the values in the Handerson equation:

pH=4.20+log(\frac{0.025}{0.05})

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by A^- .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of A^- from part a = 0.025(100) = 2.5

mili moles of HCl that is H^+ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

A^-+H^+\rightarrow HA

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

pH=4.20+log(\frac{1.5}{6})

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or OH^- added to the original buffer = 0.05(15.0) = 0.75

This OH^- reacts with HA to form A^- as:

HA+OH^-\rightarrow H_2O+A^-

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of A^- after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

pH=4.20+log(\frac{3.25}{4.25})

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0
2 years ago
A 0.080 kg robin sits on a perch 6.0 meters above the ground. What is the potential energy of the robin
Nesterboy [21]

Answer:

4.704J

Explanation:

The following data were obtained from the question:

m = 0.080kg

h = 6.0m

g = 9.8m/s^2

P.E =?

P.E = mgh

P.E = 0.08 x 9.8 x 6

P.E = 4.704J

Therefore, the potential energy of the robin is 4.704J

6 0
2 years ago
Calculate the molality of 25.0 grams of potassium bromide dissolved in 900 mL of pure water. (density of pure water is 1.0 g/mL)
Orlov [11]

Answer:

0.23mol/kg

Explanation:

Given parameters:

Mass of Potassium Bromide  = 25g

Volume of pure water = 900mL

Unknown:

Molality = ?

Solution:

Molality is one of the ways of measuring concentration. It is the number of moles of solute in given mass of solvent.

 Mathematically;

        Molality  = \frac{number of moles }{mass of solvent}

Number of moles of KBr;

   Number of moles  = \frac{mass}{molar mass}  

      Molar mass of KBr = 39 + 80  = 119g/mol

 Number of moles  = \frac{25}{119}   = 0.21mole

Mass of solvent;

   Mass of solvent  = density of water x volume of water

   Mass of solvent = 1.0g/mL x 900mL = 900g  = 0.9kg

Now;

 Molality  = \frac{0.21}{0.9}   = 0.23mol/kg

8 0
2 years ago
Balance the following KMnO4 + HCl = KCl + MnCl2 +H2O + Cl2​
Vadim26 [7]
2KMno4+16hcl=2kcl+2mncl2+8h2o+5cl2
6 0
3 years ago
PLEASE ANSWER FAST!!!!!!
lara31 [8.8K]

Answer:

A  by 0,5 cubic centimeters

Explanation:

If you use the density equation:

d=\frac{mass}{volume}→v=\frac{m}{d}

For A:

v=\frac{9[g]}{10g/cm^3} =0,9[cm^3]

For B:

v=\frac{4[g]}{10g/cm^3} =0,4[cm^3]

Doing the difference A-B=0,5 cubic centimeter

3 0
3 years ago
Read 2 more answers
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