Answer:
See the answer and explanation below, please.
Explanation:
The formulas and names of the compunds are:
Na3 P04 (Sodium Phosphate )-->The phosphate suffix corresponds to the highest number of phosphorus oxidation (+5)
Zn PO4 (Zinc Sulfate)-->The phosphate suffix corresponds to the highest number of phosphorus oxidation (+5)
KN03 ( Potassium Nitrate)-->The nitratete suffix corresponds to the highest number of nytrogen oxidation (+5)
Fe 2 (C03)3 (Iron (III) Carbonate / Dihydrous Trioxide / Ferric Carbonate)-->The carbonate suffix corresponds to the highest number of carbon oxidation (+4)
Pb (Co2)2 (Lead carbonate (IV) / Lead trioxocarbonate (IV)))-->The carbonate suffix corresponds to the highest number of carbon oxidation (+4)
91.4 grams
91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
C = mol/volume
2.45M=mol/0.5L
2.45M⋅0.5L = mol
mol = 1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225mol * 
mol=1.225
=1.225 mol . 
=1.225 . 74.6
=91.4g
therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
What is 1 molar solution?
In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.
58.44 g make up a 1M solution of NaCl.
To learn more about molar solution visit:
brainly.com/question/10053901
#SPJ4
Im not sure what the answer is unless you put the options up for the multiple choice
Answer: -
The mechanism for Br production from NBS is shown in the attached file.
N-bromo succinimide is a reagent used to brominate. It is most commonly used in allylic bromination, where a hydrogen on a carbon adjacent to a double bond is replaced.
It is selective in nature, unaffecting the normal double bond as compared to only bromine.
It also has a low concentration of bromine as an added advantage.
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>
Where Q is the amount of energy transferred (J), m is
the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
According to the given data,
Q = 300 J
m = 267 g
<span>
c = ?
ΔT = 12 °C</span>
By applying the
formula,
<span>300 J = 267 g x c x
12 °C
c = 0.0936 J g</span>⁻¹ °C⁻<span>¹
Hence, specific heat of the given substance is </span>0.0936 J g⁻¹ °C⁻¹.
