Best* and are there answer choic
<span>Ca has 2 valence electrons and F has 7, they would end up having to share electrons but Ca or F has to keep one because in total there are 9 valence elections and only 8 are needed to make this pair "happy". that's all i can provide. i hope it helped some</span>
Answer:
Option E. Zirconium
Explanation:
From the question given above, the following data were obtained:
Length of side (L) of cube = 0.2 cm
Mass (m) of cube = 52 mg
Name of the unknown metal =?
Next, we shall determine the volume of the cube. This can be obtained as follow:
Length of side (L) of cube = 0.2 cm
Volume (V) of the cube =?
V = L³
V = 0.2³
V = 0.008 cm³
Next, we shall convert 52 mg to g. This can be obtained as follow:
1000 mg = 1 g
Therefore,
52 mg = 52 mg × 1 g / 1000 mg
52 mg = 0.052 g
Thus, 52 mg is equivalent to 0.052 g.
Next, we shall determine the density of the unknown metal. This can be obtained as follow:
Mass = 0.052 g.
Volume = 0.008 cm³
Density =?
Density = mass / volume
Density = 0.052 / 0.008
Density of the unknown metal = 6.5 g/cm³
Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium
Answer: New pressure inside the flask would be 148.8 kPa.
Explanation: The combined gas law equation is given by:
As the flask is a closed flask, so the volume remains constant. Temperature is constant also.
So, the relation between pressure and number of moles becomes
- Final conditions: When additional 3 puffs of air is added
Putting the values, in above equation, we get
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
