Answers
Final answer:
An example of a solution is salt and water.
Explanation:
An example of a solution is salt and water. When salt is mixed with water, it dissolves and forms a homogeneous mixture where the salt particles are evenly distributed throughout the water.
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cual es su reacio
Answers
Answer:
corbonization for dictionnal
Explanation:
correct
Answer:
neutralization reaction forms water and salt , neutralization is the reaction!
Explanation:
B) increasing the temperature of the reactant.
C) adding a catalyst to the reaction.
D)adding more reactant
Answers
Answer:
C)
Explanation:
C) adding a catalyst to the reaction.
Answers
The number of protons neutrons, and electrons in each of the following species given are below;
What is the atomic number?
The total number of protons present in an atom is known as the atomic number of that atom. The atomic number has no correlation either with the number of neutrons or the number of electrons present inside an atom.
15N7 ⇒ 7 electrons, 8 neutrons, 7 protons
33S16 ⇒ 16 protons, 16 electrons, 17 neutrons
63Cu29 ⇒ 29 electrons, 34 neutrons,29 protons
84Sr38 ⇒ 38 electrons, 46 neutrons,38 protons
130Ba56 ⇒ 56 electrons, 74 neutrons,56 protons
186W74⇒ 74 electrons, 112 neutrons,74 protons
202Hg80 ⇒ 80 electrons, 122 neutrons ,80 protons
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Answer:
1. 7 protons, 7 electrons, 8 neutrons
2. 16 protons, 16 electrons, 17 neutrons
3. 29 protons, 29 electrons, 34 neutrons
4. 38 protons, 38 electrons, 46 neutrons
5. 56 protons, 56 electrons, 74 neutrons
6. 74 protons, 74 electrons, 112 neutrons
7. 80 protons, 80 electrons, 122 neutrons
Answers
0.0340 g O2
Step 1. Write the balanced chemical equation
4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3
Step 2. Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]
= 4.250 mmol Fe^(2+)
Step 3. Calculate the moles of O2
Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]
= 1.062 mmol O2
Step 4. Calculate the mass of O2
Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2
= 0.0340 g O2
0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.
Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol
According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:
Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2
Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):
Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2
Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
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Answers
Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
The empirical formula of the compound is PSCl₃.
To find the empirical formula, we first need to find the moles of each element in the compound. We can do this by dividing the mass of each element by its molar mass. The molar masses of the elements are:
P = 30.97 g/mol
S = 32.06 g/mol
Cl = 35.45 g/mol
The mass percentages given are for 100 g of the compound. So, the mass of each element in 100 g of the compound is:
P = 18.28 g
S = 18.93 g
Cl = 62.78 g
The moles of each element are then:
P = 18.28 g / 30.97 g/mol = 0.590 mol
S = 18.93 g / 32.06 g/mol = 0.590 mol
Cl = 62.78 g / 35.45 g/mol = 1.770 mol
The smallest whole number ratio of the moles of each element is 1:1:3. So, the empirical formula of the compound is PSCl3.
The molecular formula of the compound can be the same as the empirical formula, or it can be a multiple of the empirical formula. The molecular formula is not given, so we cannot say for sure what it is. However, we can say that the molecular formula must be a whole number multiple of the empirical formula PSCl3.
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