Answer:
Explanation:
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In this case, since the density in the international system of units is given in terms of kg for the mass and L for the volume, we need to perform a process of units conversions from mg and dL to kg and L as show below:
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Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
Answer:
Acid - A compound that increases hydrogen ions (H+) when it is dissolved in a solution
pH - A value from 0 to 14 that is used to specify how acidic or basic a compound is when it is dissolved in water
Base - A compound that increases hydroxide ions (OH−) when it is dissolved in a solution
Litmus paper - Used to measure the pH of substances by determining their hydrogen ion concentration
We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
Answer:
[Ar] 4s²
Explanation:
Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.
The full electronic configuration for calcium is given as;
1s²2s²2p⁶3s²3p⁶4s²
The condensed electronic configuration is given as;
[Ar] 4s²
