Conjugate base pairs are acid and bases having common features. These features are the equal gain or loss of protons of the pairs. Conjugate pairs should always be one base and one acid. One would not exist without the other. Conjugate acids are the substances that gains protons while conjugates bases are those that loses protons. <span>The substances in the equilibrium reaction that is given is identified as follows:
HCO3^- + H2O <-----> CO3^2- + H3O^+
acid base conjugate base conjugate acid
HCO3^- ion is an intermediate molecule of CO2 and CO3^2-. When we add OH- to HCO3^-, we produce CO3^2-. And when we add H+ to HCO3, we produce CO2. </span>
Answer:
The properties of the Hardness are:
Hardness is a physical property
Hardness is indicative of the strength of chemical bonds between elements.
Diamond can scratch quartz Explanation:
Answer:
See explanation
Explanation:
Using the formula
°C = (F-32) × 5/9
Where;
°C = temperature in degrees centigrade
F= temperature in Fahrenheit
F= (9/5 ×°C) +32
F= (9/5 × 110) + 32
F= 230°F
To convert -78°C to Kelvin
-78°C + 273 = 195 K
Answer : The correct option is, Malleable, shiny, and able to conduct heat or electricity.
Explanation:
Metals : Metals are the elements which can easily loose electrons and forms cations.
Properties of metals :
- They are lustrous (shine).
- They are malleable and ductile (flexible).
- They conduct heat and electricity.
- The metallic oxides are basic in nature.
- They form cations in an aqueous solution.
Non-metals : Non-metals are the elements which can easily gain electrons and form an anion.
Properties of non-metals :
- They are non-lustrous.
- They are brittle and hard in nature.
- They do not conduct heat and electricity.
- The non-metallic oxides are acidic in nature.
- They form anions in an aqueous solution.
Hence, from the given options the correct option for metal is, Malleable, shiny, and able to conduct heat or electricity.
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
