Answer:
[KBr] = 454.5 m
Explanation:
m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.
In this case, the moles of solute are 0.25 moles.
Let's determine the mass of solvent in kg.
Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.
1.1 g = mass of solvent / 0.5L, according to density.
mass of solvent = 0.5L . 1.1g/L = 0.55 g
We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg
m = mol/kg → 0.25 mol /5.5×10⁻⁴ kg = 454.5 m
Answer:
The standard enthalpy of combustion of solid urea ((CO(NH2)2) is -632 kJ mol-1 at 298 K and its standard molar entropy is 104.60 J K-1 mol-1
Explanation:
Answer:
X= Be
Y= B
Z=O
Explanation:
From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.
From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.
From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
Answer:
We can't see the options so we don't know what we can put
Explanation:
