Answer:
Explanation:
The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.
Metallicity is a measure of the tendency of atoms of an element to lose electrons.
a.
Down a periodic group, metallicity increases.
b.
Across a period from left to right electropositivity or metallicity decreases.
Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.
Answer : The amount left of leutium-176 will be, 2.10 g
Solution :
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the amount left of the sample.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant =
t = decay time =
a = initial amount of the sample = 16.8 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the amount left of leutium-176 will be, 2.10 g
Answer:
20.93 g
Explanation:
From the question given above, the following data were obtained:
Heat (Q) = 3.5 KJ
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Mass (M) =?
Next, we shall convert 3.5 KJ to J. This can be obtained as follow:
1 KJ = 1000 J
Therefore,
3.5 KJ = 3.5 KJ × 1000 J / 1 KJ
3.5 KJ = 3500 J
Next, we shall determine the change in the temperature of the water. This is illustrated:
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 66 – 26
ΔT = 40 °C
Finally, we shall determine the mass of the water. This can be obtained as follow:
Heat (Q) = 3500 J
Change in temperature (ΔT) = 40 °C
Specific heat capacity (C) = 4.18 J/gºC
Mass (M) =?
Q = MCΔT
3500 = M × 4.18 × 40
3500 = M × 167.2
Divide both side by 167.2
M = 3500 / 167.2
M = 20.93 g
Therefore, the mass of the water is 20.93 g
There are two kinds of mixtures
a) homogeneous : the boundary of the two components is not physically distinct
b) heterogeneous:the boundary of the two components is physically distinct
the following separation techniques are common for mixtures
1) filtration: if the two components are forming heterogeneous mixture we can separate them by filtration.
2) boiling: if boiling point of one of the components is less than other
3) magnetic separation: if one of the component is magnetic
4)sieve method: for solid components with difference in size of particles
5) hand picking
Thus the correct match will be as shown in the figure