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The completed question is given below
Different microscopic organisms cause different diseases. When a dangerous microscopic organism is transferred from one person to many people, the disease it carries will spread. What are the mode of transmission of infectious diseases
Explanation:
Infectious diseases are the diseases that are transmissible from one person to another, most of these diseases are usually transmitted through air when the infected person sneezes or cough.
Another means of transmission of diseases is by contact, this usually happens when the infected person come in contact with another person.
Answer:
Explanation:
Q1. Mass of Cu
(a) Write the equation for the half-reaction.
Cu²⁺ + 2e⁻ ⟶ Cu
The number of electrons transferred (z) is 2 mol per mole of Cu.
(b) Calculate the number of coulombs
q = It
(c) Mass of Cu
We can summarize Faraday's laws of electrolysis as
Note: The answer can have only two significant figures because that is all you gave for the time.
Q2. Current used
(a) Write the equation for the half-reaction.
Ag⁺ + e⁻ ⟶ Ag
The number of electrons transferred (z) is 1 mol per mole of Ag.
(a) Calculate q
(b) Calculate the current
t = 3 h = 3 × 3600 s = 10 800 s
Note: The answer can have only one significant figure because that is all you gave for the time.
Answer:
The mass of third isotope is 25.98 amu.
Explanation:
Mass of Mg²⁴ = 23.985 amu
Mass of Mg²⁵ = 24.986 amu
Abundance of Mg²⁴ = 78.99 %
Abundance of Mg²⁵ = 10%
Abundance of third isotope = 11.01%
Atomic weight of magnesium = 24.305
Solution:
Atomic mass = ( % age of first isotope × its atomic mass) + (% age abundance of second isotope × its atomic mass) + ( % age of third isotope × its atomic mass ) / 100
Now we will put the values in formula.
24.305 = (78.99 × 23.985 ) + (10 ×24.986 ) + ( 11.01 × X) / 100
24.305 = 1894.56 + 249.9 +( 11.01 × X ) / 100
24.305 × 100 = 2144.46 + ( 11.01 × X)
2430.5 - 2144.46 = (11.01 × X)
286.04 / 11.01 = X
25.98 = X
The mass of third isotope is 25.98 amu.
Answer:Boiling point of a solution is found to be 100.34
o
C. Boiling point of pure
water is 100
o
C.
The elevation in the boiling point ΔT
b
=100.34−100=0.34
o
C.
12 gm glucose (molecular weight 180 g/mol) is dissolved in 100 gm water.
The number of moles of glucose =
180g/mol
12g
=0.0667mol
Mass of water =100g×
1000g
1kg
=0.100kg
Molality of solution m=
0.100kg
0.0667mol
=0.667mol/kg
The elevation in the boiling point ΔT
b
=K
b
×m
0.34
o
C=K
b
×0.667mol/kg
K
b
=0.51
o
Ckg/mol
The molal elevation constant for water is 0.51
o
Ckg/mol.
Explanation: