Answer:
Explanation:
The relative massive alpha particles could go through the gold foil without being deviated of their trajectory or only small deviations due to the uniformity distribution positive charge of the protons.
Answer:
6. d, 7. a
Explanation:
6. Molarity is a number of moles solute in 1 L solution.
7. 1 L solution - 2.5 mol K2CO3
20 L - x mol K2CO3
x =20*2.5/1 = 50 mol K2CO3
Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol
99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.
Actually KCO3 does not exist, in reality it should be K2CO3.
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
The 3 indicates the third electron shell. (Which has only 1 electron in it in this configuration)
Hope this helps! :)
Answer:
group 8 of the periodic table
Explanation:
