A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylinder after the borax solution had cooled to a certain temperature T. The student rinses the sample into a small beaker using distilled water, and then titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCl solution is needed to reach the endpoint of the titration.Calculate the value of Ksp for borax at temperature T.
Answers
Answer:
ksp = 0,176
Explanation:
The borax (Na₂borate) in water is in equilibrium, thus:
Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)
When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)
The ksp is defined as:
ksp = [borate²⁻] [Na⁺]²
Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:
B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻
The moles of HCl that reacts with B₄O₇²⁻ are:
0,500M×0,01200L = 6,00x10⁻³ mol of HCl
As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:
6,00x10⁻³ mol of HCl× = 3,00x10⁻³ mol of B₄O₇²⁻
For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺
The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M
And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M
Replacing in the expression of ksp:
ksp = [0,353] [0,706]²
ksp = 0,176
I hope it helps!
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Answers
Half-life of a radioactive element
The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay
Calculating the original amount of U-238
- The number of moles present in each element is first determined:
- Number of moles = mass/molar mass
For Pb-206:
mass = 0.37 mg = 0.37 * 10⁻³ g
molar mass = 206 g/mol
number of moles = 0.37 * 10⁻³ g/206 g/mol
number of moles of Pb-206 = 1.79 * 10⁻⁶ moles
For U-238:
mass = 0.95 mg = 0.95 * 10⁻³ g
molar mass = 238 g/mol
number of moles = 0.95 * 10⁻³ g/238 g/mol
number of moles = 3.99 * 10⁻⁶ moles
Assuming that all the Pb-206 were formed from U-238
- Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208
Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles
Initial moles of U-238 = 5.78 * 10⁻⁶ moles
One mole of U-238 contains = 6.02 * 10²³ atoms
5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms
Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms
Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms
Learn more about Half-life at: brainly.com/question/4702752
b. 0.03954 g to 0.040 g
c. 20.0332 g to 20,0 g
d. 04.05438 g to 4,054 g
e. 103.692 g to 103.7g
Answers
Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
Which of the following examples illustrates a number that is correctly rounded to three significant figures?
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
Answers
Answers
Explanation:
..(1)
Manganese (II) carbonate and oxygen reacts to give manganses (III) oxide with carbon-dioxide gas.
..(2)
Manganese (III) oxide reacts with aluminum metal to give aluminum oxide and manganese metal
On adding (1) and 2(2) we will get the net net chemical equation for the production of manganese. So,
The net balanced equation we got:
Final answer:
The production of manganese from manganese (II) carbonate, oxygen, and aluminum involves two chemical reactions. Heating manganese (II) carbonate with oxygen yields manganese(III) oxide and carbon dioxide. This manganese (III) oxide then reacts with aluminum to produce manganese and aluminum oxide.
Explanation:
The chemical reaction can be described in terms of a net chemical equation after properly balancing the chemicals involved on both sides of the equation. The first step is to understand and correctly write down the formula of reactants and products for the reaction. Here: Manganese (II) Carbonate (MnCO3), Oxygen (O2), and Aluminum (Al).
In the process of manganese production, Manganese (II) Carbonate is first heated in the presence of oxygen to yield Manganese (III) Oxide (Mn2O3) and Carbon Dioxide (CO2):
MnCO3 + 1/2 O2 -> Mn2O3 + CO2.
The resulting Manganese (III) Oxide reacts with Aluminum (which acts as a reducing agent in this case) to generate Manganese and Aluminum Oxide:
3 Mn2O3 + 4 Al -> 6 Mn + 2 Al2O3.
These two equations combined represent the net chemical equation for the production of manganese from manganese (ii) carbonate, oxygen, and aluminum.
Learn more about Chemical Reactions here:
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Answers
Answer:
less, decreases
Explanation:
When the pressure of an atmosphere occurs because of the force exerted so at the time of the higher altitudes, the air mass i.e. above the earth should be less as the air is attracted towards surface of an earth because of the gravity and air contains the mass that shows near the surface area so automatically the air density reduced due to which the mass also decreased
Answers
Answer:
Explanation:
the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole
106*0.251=26.606 grames