Answer:
I = 1.23 A
Explanation:
Given that,
The resistance of the lightbulb, R = 96.8 Ω
Voltage, V = 120 V
We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.
So, the current flows through the bulb is 1.23 A.
Answer :
(a) The overall equation is:
(b) The intermediates are :
Explanation :
<u>Part (a) :</u>
(1) (fast)
(2) (slow)
(3) (fast)
By adding the three equations and cancelling the common terms on both side, we will get the overall equation.
<u>Part (b) :</u>
Intermediates are generated and consumed in the mechanism and do not include in the overall equation.
Since, intermediates will not include in the overall mechanism.
The intermediates are :
Answer:
Selenium
Explanation:
Oxygen is present in the group sixteen.
This group consist of oxygen, sulfur, selenium, tellurium and polonium.
All of these have same number of valance electrons thus they show similar properties.
From the given list of elements only selenium is belong to oxygen family thus it has properties similar to the oxygen.
Electronic configuration of oxygen:
O = [He] 2s² 2p⁴
Electronic configuration of selenium:
Se = [Ar] 3d¹⁰ 4s² 4p⁴
While aluminium, boron and phosphorus are belongs to other groups that's why their properties are not similar to the phosphorus.
phosphorus is present in group 15.
Boron and aluminium are present in group 13.
<h3>
Answer:</h3>
132.03 g
<h3>
Explanation:</h3>
<u>We are given;</u>
- The equation for the reaction as;
Fe₂O₃ + 3CO → 2Fe + 3CO₂
- Molar masses of CO and CO₂ as 28.01 g/mol and 44.01 g/mol respectively
- Mass of CO as 84 grams
We are required to calculate the mass of CO₂ that will produced.
<h3>Step 1: Calculate the number of moles of CO</h3>
Moles = Mass ÷ Molar mass
Molar mass of CO = 28.01 g/mol
Therefore;
Moles of CO = 84 g ÷ 28.01 g/mol
= 2.9989 moles
= 3.0 moles
<h3>Step 2: Calculate the number of moles of CO₂</h3>
- From the reaction, 3 moles of CO reacts to produce 3 moles of CO₂
- Therefore; the mole ratio of CO to CO₂ is 1 : 1
- Hence; Moles of CO = Moles of CO₂
Moles of CO₂ = 3.0 Moles
But; mass = Moles × molar mass
Thus, mass of CO₂ = 3.0 moles × 44.01 g/mol
= 132.03 g
Hence, the mass of CO₂ produced from the reaction is 132.03 g
1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.
2) Equation:
2Na OH + H2SO4 --> Na2 SO4 + 2H2O
3) molar ratios
2 mol NaOH : 1 mol H2SO4
4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution
M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4
5) Number of moles of NaOH
2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH
6) Concentration of the solution of NaOH
M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M
7) Standardize the solution of HCl
Chemical reaction:
NaOH + HCl --> NaCl + H2O
8) Molar ratios
1 mol NaOH : 1 mol HCl
9) Number of moles of NaOH in 27.5 ml
M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH
10) Number of moles of HCl
1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl
11) Concentration of the solution of HCl
M = n / V = 0.01169 mol / 0.100 l = 0.1169 M
Rounded to 3 significant figures = 0.117 M
Answers:
[NaOH] = 0.425 M
[HCl] = 0.117 M