The conclusive test for iron(II) ions is the test by the use of potassium hexacyanoferrate III solution.
In qualitative analysis certain reagents are used to test for the presence of certain cations or anions. Those reagents react in a certain way with those reagents. Usually, a positive test may involve a color change, formation of a precipitate or evolution of a gas.
In the case of iron(II) ions, potassium hexacyanoferrate III solution is used in the conclusive qualitative test for the ion. A positive test involves the appearance of a deep blue precipitate.
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Answer:
please brainlist answer
Explanation:
The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.
Answer:
Explanation:
Hello,
In this case, for the given chemical reaction, we can notice there is a 4:2 molar ratio between the burned moles of gallium and the yielded moles of gallium sulfide, therefore, we compute them as shown below:
Then, by using the molar mass of gallium sulfide (235.64 g/mol), we directly compute the grams:
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Here's the answer, I remember doing this problem last year.
23.5 degrees north, 77 degrees west
Answer:
Part one: B. Kc decreases
Part two: B. Is equal to Kc
Part three: B. Run in the reverse direction to reestablish equilibrium
Part four: A. Increase
Explanation:
Part one: Sulfur dioxide combines with oxygen to form sulphur trioxide in an exothermic reaction. If the temperature is suddenly is increased, while the reaction is at equilibrium, the backward reaction (the endothermic one) is favored to "sweep up the excess heat". An increase in reactants means a decrease in Kc since the denominator(reactants) is becoming bigger while the numerator (products) become smaller.
Part two: Qc is a varying version of Kc. For this set of circumstances, it will be equal to Kc since Kc varies with temperature
Part three: The reaction must run in the reverse to reestablish the equilibrium.
Part four: The concentration of of oxygen will increase as more of the reactants are formed
The increase in temperature for this exothermic reaction will cause the value of Kc to decrease, the value of Qc to be greater than Kc, the reaction to run in the reverse direction, and the concentration of O2 to increase.
The given chemical reaction represents a type of equilibrium reaction, specifically an exothermic reaction, as it produces sulfur trioxide (SO3), which releases heat. According to Le Chatelier's principle, to maintain equilibrium, if a system is disturbed by an external factor, the system will adjust accordingly.
Here are my answers to the specific questions:
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___2.Variables
___3.Conclusion
___4.Scientific Method
___5.Procedure
a.The steps you take to complete the experiment
b.Factors that changes in an experiment
c.A possible solution to a problem
d.The result of the experiment
e.The process scientist follow to complete an investigation
Answer:
1. d
2. b
3. d
4. e
5. a
explanation:
there's nothing else to explain
Answer:
0.190 M
Explanation:
Let's consider the neutralization reaction between HCl and NaOH.
HCl + NaOH = NaCl + H2O
11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:
0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol
The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.
1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:
M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M
Answer:
The initial concentration of HCl was 0.1904 M
Explanation:
Step 1: Data given
Volume of HCl solution = 10.0 mL = 0.010 L
Volume of a NaOH solution = 11.9 mL = 0.0119 L
Molarity of NaOH solution = 0.160 M
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate the concentration of HCl
C1*V1 = C2*V2
⇒with C1 = the concentration HCl = TO BE DETERMINED
⇒with V1 = the volume of HCl = 0.010 L
⇒with C2 = the concentration of NaOH = 0.160 M
⇒with V2 = the volume of NaOH = 0.0119 L
C1 * 0.010 L = 0.160 M * 0.0119 L
C1 = (0.160 M * 0.0119 L) / 0.010 L
C1 = 0.1904 M
The initial concentration of HCl was 0.1904 M
Answer:
23.84g CH30H
32.81g H2O
Explanation:
We will be using the definition of mole fraction to determine the relationship between the number of moles of methanol,
CH3OH , and the number of moles of water.
But mole fraction gives the ratio between the number of moles of a component i of a solution to the total number of moles present in that solution.
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
In a 66.0g aqueous solution of methanol with a mole fraction of 0.290, the mass of the methanol is approximately 19.14g and the mass of the water is approximately 46.86g.
In this aqueous solution of methanol (CH4O), we know that its mass is 66.0g and the mole fraction of methanol is 0.290. The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution.
In order to find the mass of each component, namely the methanol and the water, we first need to establish that if the mole fraction of methanol is 0.290, the mole fraction of water must be 0.710 (because the total of all mole fractions in a solution is always equal to 1).
We then can set up the following proportion: mass of methanol/mass of water = mole fraction of methanol/mole fraction of water. After solving this equation, the mass of methanol will be approximately 19.14g and the mass of the water will be approximately 46.86g.
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