Answer:
1. Zn is oxidize.
Cu is reduced.
2. Fe is oxidize.
H2 is reduced.
Explanation:
1. Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
From the equation,
Zn changes oxidation number from 0 to +2. Therefore Zn is oxidized.
Cu2+ changes oxidation number from +2 to 0. Therefore, Cu2+ is reduced
2. 3Fe(s)+4H2O(g)→Fe2O3(s)+4H2(g)
From the equation,
Fe changes oxidation number from 0 to +3. Therefore, Fe is oxidized.
H changes oxidation number from +1 to 0. Therefore, H is reduced
Answer:
Option-B (Carbon and Silicon)
Explanation:
Among the given pairs only carbon and silicon have the most similar properties. This is because,
Sodium and Magnesium belong to different groups. Sodium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Magnesium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.
Example:
2 Na + Cl₂ → NaCl
Mg + Cl₂ → MgCl₂
As shown in reactions when Sodium and Magnesium are treated with Cl₂ they give a products with different proportions.
Carbon and Silicon show almost same properties because both belong to Group IV hence both are capable of forming four bonds. Also, they share the same property of self linkage in making a long chains.
Argon and Chlorine also belong to two different groups. Argon is present in Group VIII (Noble Gases) and Chlorine is present in Group VII (Halogens). Hence, Argon is an inert specie which is non reactive while Chlorine gives different reaction easily.
Potassium and Calcium belong to different groups. Potassium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Calcium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.
Example:
2 K + Cl₂ → KCl
Ca + Cl₂ → CaCl₂
As shown in reactions when Potassium and Calcium are treated with Cl₂ they give a products with different proportions.
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
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- A car travels 120km due to North .
- At last the car reached on town B from town A .
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- The car's total displacement from town A to B .
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✍️ See the attachment diagram .
✯ Hence, Total displacement from town A to town B is “ <u>8</u><u>2</u><u>k</u><u>m</u> ” .
1) Chemical equation
C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O
2) Molar ratios
1 mol C2H5OH : 3 mol O2 : 2 mol CO2 : 3 mol H2O
3) Amount of ethanol burned
D = M / V => M = D * V = 0.789 g/ml * 4.61 ml = 3.637 g
Number of moles = mass in grams / molar mass
molar mass of C2H5OH = 2 *12 g/mol + 6*1.0 g/mol + 16.0 g/mol = 46.0 g/mol
=> number of moles of C2H5OH = 3.637 g / 46.0 g/mol = 0.079 mol
4) Amount of oxygen, O2
number of moles of O2 = mass in grams / molar mass
number of moles O2 = 15.70g / 32.0 g/mol = 0.49 mol
5) Limiting reactant
Theoretical ratio: 1 mol C2H5OH / 3 mol O2
Actual ratio: 0.079 mol / 0.49 mol = 0.16
=> there is more oxygen than needed to burn all the ethanol => ethanol burns completely and it is the limiting reactant.
6) Theoretical yield of H2O
1 mol C2H5OH / 3 mol H2O = 0.079 mol C2H5OH / x
=> x = 0.079 mol C2H5OH * 3 mol H2O / 1 mol C2H5OH
=> x = 0.237 mol H2O
Conversion to grams:
mass = molar mass * number of moles
mass H2O = 18.0 g/mol * 0.237 g/mol = 4.27 g <---- theoretical yield
7) grams of H2O collected (yield)
M = V * D = 3.27 ml * 1.00 g/ml = 3.27 g <----- actual yield
8) Percent yield
Percent yield = (actual yield / theoretical yield) * 100 = (3.27 g / 4.27) * 100 = 76.6%
Answer: 76.6%