I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.
The density of water is 1kg/L. Since the density of the block is less, it will float.
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
<span>0.48 grams.
Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced.
To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So
0.0075 mol * 63.546 g/mol = 0.476595 g.
Round the results to 2 significant figures, giving 0.48 grams.</span>
Balanced Half reactions are:
At anode 2
==> Cl₂+ 
+ H₂O ==> 
+ 2
+
At Cathode: 2 + ==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, , 
and
Now at Anode reaction will occur as given:
2 ==> Cl₂+ + H₂O ==> + 2 + (will occur)
At Cathode:
2 + ==> H₂ (will occur)
At Cathode:
+ ==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, will be reduced before .
The reduction potentials for is zero whereas reduction potential for is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
To learn more about the half reaction please click on the link brainly.com/question/13186640
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