The density of Ca will be between that of Mg and Sr
Explanation:
Ca, Mg and Sr are group II elements. They are called alkali earth metals. The correct order of the elements in this group are: Be, Mg, Ca, Sr, Ba and Ra.
Density is an intensive property of matter which describes the amount of matter(mass) per volume of a substance.
- Density varies proportionally with mass. The higher the mass, the higher its density.
- On the periodic table, atomic mass which the number of protons and neutrons in the nucleus of an atom increases down the group.
- This implies a progradation in the value of density down the group. Therefore one expects that the value of density of Ca will fall between that of Mg and Sr. It cannot be more than 2.6g/cm³ nor less than 1.74g/cm³.
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HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
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This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
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In seawater, salt is the solute and water is the solvent.
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