Solved: An Unknown Element Contains 35 Protons, 36 Electro... | ...
Answer:
- <u><em>Sodium chloride</em></u>
Explanation:
The attached graph with a green and a red arrow facilitates the understanding of this explanation.
To read the <em>solubility </em>on the <em>graph</em>, you can start with the temperature, on the x-axis.
The red vertical arrow shows how, departing from the <em>40ºC temperature</em> on the x-axis, you intersect the<em> solutibility curve </em>of sodium chloride at a height (y-axis) corresponding to <em>60 g/100cm³ of water</em> (follow the green horizontal arrow).
Hence, <em>sodium chloride is the salt that can dissolve at a concentration of about 60g/100cm³ of water at 40ºC.</em>
For the first Question,
d. It is a substance that cannot be chemically broken down any further.
For the second Question,
Please right the whole options completely so that I can answer
Answer:
-1960 kJ.
Step-by-step explanation:
C₄H₄O₄(s) + 3O₂(g) ⟶ 2H₂O((ℓ)) + 4CO₂(g) + Energy
There are three energy flows in this reaction.
From combustion + warm water + warm calorimeter = 0
q₁ + q₂ + q₃ = 0
nΔH + mCwΔT + CcalΔT = 0
<em>Data:
</em>
Mass of fumaric acid = 1.1070 g
Mass of water = 1.093 × 10³ g
Cw = 4.184 J·°C⁻¹g⁻¹
T₁ = 21.10 °C
T₂ = 24.52 °C
Ccal = 891.1 J·°C⁻¹
Calculations:
(a) <em>q₁
</em>
n = 1.1070 g × (1 mol/116.07 g)
n = 0.009 537 mol
q₁ = 0.009 537ΔH J
(b) <em>q₂
</em>
ΔT = 24.52 – 21.10
ΔT = 3.42°C
q₂ = 1093 × 4.184 × 3.42
q₂ = 15 640 J
(c) <em>q₃
</em>
q₃ = 891.1 × 3.42
q₃ =3048 J
(d) <em>ΔH</em>
0.009 537ΔH + 15 640 + 3048 = 0
0.009 537ΔH + 18 688 = 0
0.009 537ΔH = -18 688
ΔH = -18 688/0.009 537
ΔH = -1 959 413 J/mol
ΔH = -1960 kJ/mol
This is quite different from the actual value of -1334.70 kJ·mol⁻¹
Answer:
0.081g
Explanation:
first step:calculate the molar conc of the stock solution
molar conc=mass/molar mass ×1000/vol(ml)
molar mass of Co(N03)2=182.943
molar conc=3g/182.943 ×1000/100ml
=0.163M
To calculate the conc of the resulting solution we apply the dilution principle
n=cv
C1V1=C2V2
C1=conc of stock solution=0.163M
V1=4mL
V2=275ml(final volume)
by making 'C2" the subject C2=0.163×4/275
=0.0024M
by applying the formular for calculating molar conc of a solution,we can calculate the mass of CO(NO3)2 in the reulting solution
molar conc=mass/molar mass ×1000(L)/vol(mL)
mass=0.0024×182.943×275/1000
=0.1193g
Mass of CO(NO3)2 in the resulting solution=0.1199g
since CO+(NO3)2=CO(NO3)2
Percentage by mass of (NO3)2 in CO(NO3)2=(NO3)2/CO(NO3)2
=124/182.943×100
=67.8%
it means 67.8% of 0.1193g is the mass of (NO3)2=0.678×0.1199
=0.081g