Answers
Answer:
c) 0.080 M Al₂(SO₄)₃
Explanation:
Ion [SO₄²⁻] concentration of each solution is:
a) 0.075 M H₂SO₄: [SO₄²⁻] = 0.075M. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻
b) 0.15 M Na₂SO₄: [SO₄²⁻] = 0.15M. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻
c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 = 0.240M. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.
Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃
Related Questions
What is the question mark
A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.
Hich nuclide is most likely to undergo beta decay?which nuclide is most likely to undergo beta decay?co−52ar−35si−22mg−28
What is the difference between solid, liquid and ice? Use ice, water and steam as examples.
Answers
step by step explanation
simple distillation can be used when the temperature difference between the boiling points of two miscible liquid is at least 25°c. the temperature difference between the boiling points of kerosene and petrol is 25c. hence, this mixture can separated using simple distillation.
answer:
simple distillation
b. 1.50 x 1023 atoms Mg
c. 4.50 x 1012 atoms Cl
d. 8.42 x 1018 atoms Br
e. 25 atoms W
f. 1 atom Au
Answers
The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.
The mass in grams of 1.50 x 10²³ atoms of Mg is 5.98 g.
The mass in grams of 4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.
The mass in grams of 8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.
The mass in grams of 25 atoms of W is 3.1 x 10⁻²¹ g.
The mass in grams of 1 atom of Au is 3.27 x 10⁻²² g.
What is the mass in grams of 3.011 x 10²³ atoms F?
The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;
6.023 x 10²³ atoms = 19 g of F
3.011 x 10²³ atoms F = ?
= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)
= 9.5 g
The mass in grams of 1.50 x 10²³ atoms of Mg is calculated as follows;
6.023 x 10²³ atoms = 24g of Mg
1.5 x 10²³ atoms F = ?
= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)
= 5.98 g
The mass in grams of 4.50 x 10¹² atoms of Cl is calculated as follows;
6.023 x 10²³ atoms = 35.5 g of Cl
4.5 x 10²³ atoms Cl = ?
= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)
= 2.65 x 10⁻¹⁰ g
The mass in grams of 8.42 x 10¹⁸ atoms of Br is calculated as follows;
6.023 x 10²³ atoms = 80 g of Br
8.42 x 10¹⁸ atoms Br = ?
= (8.42 x 10¹⁸ x 80 g)/(6.023 x 10²³)
= 1.12 x 10⁻³ g
The mass in grams of 25 atoms of W is calculated as follows;
6.023 x 10²³ atoms = 74 g of W
25 atoms W = ?
= (25 x 74 g)/(6.023 x 10²³)
= 3.1 x 10⁻²¹ g
The mass in grams of 1 atom of Au is calculated as follows;
6.023 x 10²³ atoms = 197 g of Au
1 atom of Au = ?
= (1 x 197 g)/(6.023 x 10²³)
= 3.27 x 10⁻²² g
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Final answer:
This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.
Explanation:
The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.
- For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
- For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
- For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
- For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
- For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
- For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.
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Answers
In the plum pudding model, alpha particles were expected to pass through the foil at different angles and some of them reflected. They realized that there are positive charge atom because it repels. So the plum pudding was replaced by the nuclear model of atom.
An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:
Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)
Calculate the cell potential under these nonstandard concentrations.
Express the cell potential to two decimal places and include the appropriate units.
Answers
Answer:
Cell potential under non standard concentration is 4.09 v
Explanation:
To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).
Nernst equation at 25ºC is:
where
E: Cell potential (non standard conditions)
= Cell potential (standard conditions)
n: Number of electrons transfered in the redox reaction
Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)
For example, consider the following general chemical reaction:
aA + bB --> cC + dD
where
a, b, c, d: coefficient of balanced chemical reaction
A,B,C,D: chemical compounds in the reaction.
Using the previous general reaction, expression of Q is:
Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: , n and Q
Let´s calcule potential in nomal conditions ():
1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:
Half-reactions: Eo (v):
+ e- -->
-0.23
+ 2 e- +0.99
Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:
Half-reactions: Eo (v):
2 * [ + e- -->
] -0.23
1 * [ + 2 e- ] +0.99
------------------------------------------------------------------- -------------
2 + 2e- + Ni --> 2
+
+ 2e- 0.76 v
Then global balanced chemical reaction is:
2 + Ni --> 2
+
and the potential in nomal conditions is:
= 0.76 v
Also from the balanced reaction, we got number of electons transfered:
n = 2
2. Calculate Q:
Now using previous information, we can establish Q expression and we can calculate its value:
From the exercise we know:
3. Use Nernst equation:
Finally, we replace all these results in the Nernst equation:
Cell potential under non standard concentration is 4.09 v
Final answer:
To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.
Explanation:
For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.
Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.
Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).
After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.
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Answers
Answer: Option (2) is the correct answer.
Explanation:
Atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. So, it contains only 2 orbitals which are closer to the nucleus of the atom.
As a result, the valence electrons are pulled closer by the nucleus of oxygen atom due to which there occurs a decrease in atomic size of the atom.
Whereas atomic number of sulfur is 16 and its electronic distribution is 2, 8, 6. As there are more number of orbitals present in a sulfur atom so, the valence electrons are away from the nucleus of the atom.
Hence, there is less force of attraction between nucleus of sulfur atom and its valence electrons due to which size of sulfur atom is larger than the size of oxygen atom.
Thus, we can conclude that the oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
Final answer:
The oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
Explanation:
The correct option is (2) the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
To understand why the oxygen atom is smaller than the sulfur atom, we need to consider their electron configurations. Oxygen has 8 electrons and sulfur has 16 electrons. Oxygen's electron configuration is 1s²2s²2p⁴, while sulfur's electron configuration is 1s²2s²2p⁶3s²3p⁴.
The outer orbitals of an atom, which are the valence orbitals, are the ones involved in bonding. The electrons in these orbitals determine the size of the atom. In the case of oxygen and sulfur, the outer orbitals of oxygen (2p orbitals) are closer to the nucleus compared to sulfur's outer orbitals (3p orbitals). As a result, the oxygen atom is smaller than the sulfur atom.
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Answers
Answer:
See explanation below
Explanation:
In both cases the central atoms, C in CHCl₃ and O in H₂O, are sp³ hybridized .
Since they are sp³ hybridized we predict an angle between the H-C-Cl and H-O-H of 109.5 º ( tetrahedral ), but two of the sp³ orbitals in water are occupied by lone pairs.
These lone pairs do excercise more repulsion ( need more room ) than the bonds oxygen is making with hydrogen.
As a consequence of this repulsion the angles H-O-H are less than the predicted 109.5º in tetrahedra. ( Actually is 104.5 º)