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iris [78.8K]
4 years ago
10

Using the complex based titration system: 50.00 mL 0.00250 M Ca2+ titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i)

first pCa first before initiating the titration process and then (ii) at equivalence when all the Ca2+ is titrated to CaY2-. Please, use your text books and/or lecture notes to find potentially missing information about constants needed to solve the problem.
Chemistry
1 answer:
gogolik [260]4 years ago
4 0

Answer:

i) The pCa before initiating the titration is 2.6

ii) The pCa is 6.67

Explanation:

please look at the solution in the attached Word file

Download docx
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A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0
3 years ago
Magnesium has three naturally occurring isotopes with masses of 23.99 amu, 24.99 amu, and 25.98 amu and natural abundances of 78
mihalych1998 [28]

Answer:24.31

Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97

Contribution made by isotope of mass 24.99=24.99×10.00=249.9

Contribution made by isotope of mass 25.98=25.98×11.01=286.04

Total contribution=1894.97+249.9+286.04=2430.91

Average mass=2430.91÷100

=24.31

3 0
3 years ago
Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound a
Margarita [4]

Answer:

Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.

Explanation:

Alcohols belong to a group of organic compounds which contain -OH group as the functional group.

So alcohols have the same carbon -hydrogen bonds as alkanes and the        -OH functional group identifies the compound as an alcohol.

8 0
3 years ago
What is the mass of 22.4 L of H2 at STP?
Vanyuwa [196]

A. 1.01 is the right answer

Since

The formula is Pv= nRT

P=1 atm

V= 22.4 L

N= x

r= 0.0821

t = 273 k (bc it’s standard temperature)

So (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0
4 years ago
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A student mixed two chemicals to allow them to
Westkost [7]

Answer:

B .it is an exothermic reaction

Explanation:

i need brainleyest

4 0
2 years ago
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