I think it's A. because the Earth won't always have the same tilt throughout it's whole life. The tilt of the axis will change, even if it's just ever so slightly. I hope this helps and i hope i'm right :)
1. Berkelium(Berkeley, CA) 2. Dubnium(Dubna, Russia) 3. Darmstaditum (Darmstadt, Germany) 4. Erbium(Ytterby, Sweden) 5. Strontium(Strontian, Scotland) 6. Terbium(Ytterby, Sweden) 7. Yttebium(Ytterby, Sweden) 8. Yttrium(Ytterby, Sweden)
1751 - Nickel
1774 - Manganese
1781 - Molybdenum
1782 - Tellurium
1783 - Tungsten
1789 - Uranium
1789 - Zirconium
1791 - Titanium
1794 - Yttrium
1797 - Berylium
1797 - Chromium
1801 - Niobium
1802 - Tantalum
1803 - Iridium, Palladium, Rhodium
1807 - Potassium, Sodium
1808 - Boron, Barium, Calcium, Magnesium, Strontium
1814 - Cerium
1817 - Lithium, Cadmium, Selenium
1823 - Silicon
1827 - Aluminium
1828 - Thorium
1830 - Vanadium
1839 - Lanthanum
1843 - Erbium, Terbium
1844 - Ruthenium
1860 - Cesium, Rubidium
1861 - Thallium
1863 - Indium
1875 - Gallium
1878-1885 - Holmium, Thulium, Scandium, Samarium, Gadalinium, Praseodynium, Neodynium, Dysprosium
1886 - Germanium
1898 - Polonium, Radium
1899 - Actinium
1901 - Europium
1907 - Lutetium
1917 - Protactinium
1923 - Hafnium
1924 - Rhenium
1937 - Technetium
1939 - Francium
1945 - Promethium
1940-61 - Transuranium elements – (Neptunium, Plutonium, Curium, Americum, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium, Lawrencium)
Answer:
49 g/L is the concentration of the acid
Explanation:
Firstly, we proceed to write the equation of reaction.
2NaOH + H2SO4 ——-> Na2SO4 + 2H2O
We can see that 1 mole of the base reacted with two moles of the acid.
kindly note that dm^3 is same as liter
Firstly, we need to get the concentration of the reacted sulphuric acid in g/L
we use the simple titration equation below;
CaVa/CbVb = Na/Nb
From the question;
Ca = ?
Va = 25 cm^3
Cb = 20 g/L
we convert this to concentration in mol/L
Mathematically, that is concentration in g/L divided by molar mass in g/mole
molar mass of NaOH = 40 g/mol
so we have; 20g/L / 40 = 0.5 mol/L
Vb = 50 cm^3
Na = 1
Nb = 2
Where C represents concentrations, V volumes and N , number of moles
Now, substitute the values;
Ca * 25/0.5 * 50 = 1/2
25Ca/25 = 0.5
So Ca = 0.5 mol/L
Now to get the concentration of H2SO4 in g/L
What we do is to multiply the concentration in mol/L by molar mass in g/mol
That would be 0.5 * 98 = 49 g/L