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WINSTONCH [101]
2 years ago
14

Oliver needs to standardize a base solution by titration with an acid solution of known concentration. He finds the volume of th

e acid solution needed to reach the endpoint for a sample of the base solution. Arrange the steps in the order that allows Oliver to determine the concentration of the base solution.
a. Write a balanced equation for the between the analyte and the titrant.
b. Calculate the number of moles of analyte using the stoichiometric coefficients Of the equation.
c. Calculate One number moles of titrant using One volume of titrant required and the concentration of titrant
d. Calculate Ole concentration of the analyte using the number of moles of analyte and the volume of analyte titrated.
Chemistry
1 answer:
Olin [163]2 years ago
7 0
Write a balance equation for the reaction between the analyte and the titrant.
Calculate the # of moles of titrant using the volume of titrant required and the concentration of titrant.
Calculate the # of moles of analyte using the stoichiometric coefficients of the equation.
Calculate the concentration of the analyte using the number or moles of analyte and the volume of analyte titrated.
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What is the molarity of a solution prepared by dissolving 10.0g of kno3 in 250 ml of solution
maw [93]

Answer:

Mass of KNO3= 10g

Molar mass of KNO3 = 101.1032g/mol

Volume = 250ml = 0.25L

No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3

no of mole of KNO3 = 10/101.1032

No of mole of KNO3 = 0.09891

molarity of KNO3 = no of mole of KNO3/Vol (L)

Molarity = 0.09891/0.25 = 0.3956M

Molarity of KNO3 = 0.3956M

8 0
3 years ago
A certain sample of coal contains 1.60 percent sulfur by mass. when the coal is burned, the sulfur is converted to sulfur dioxid
alina1380 [7]
<span>1.40 x 10^5 kilograms of calcium oxide The reaction looks like SO2 + CaO => CaSO3 First, determine the mass of sulfur in the coal 5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4 Now lookup the atomic weights of Sulfur, Calcium, and Oxygen. Sulfur = 32.065 Calcium = 40.078 Oxygen = 15.999 Calculate the molar mass of CaO CaO = 40.078 + 15.999 = 56.077 Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight. 8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass 2.49 x 10^3 * 56.077 = 1.40 x 10^5 So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
5 0
3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
2 years ago
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0
3 years ago
How many molecules are in 97.21 grams of Sodium Chloride (NaCl)?
professor190 [17]
1.66 is the answer because it’s is
6 0
2 years ago
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