Answer:
100 g
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 400 g
Time (t) = 4 years
Half-life (t½) = 2 years
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Time (t) = 4 years
Half-life (t½) = 2 years
Number of half-lives (n) =?
n = t / t½
n = 4 / 2
n = 2
Thus, 2 half-lives has elapsed.
Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:
Original amount (N₀) = 400 g
Number of half-lives (n) = 2
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 0.25 × 400
N = 100 g
Thus, the amount of the radioactive isotope remaing is the 100 g.
Answer:
Nickel and Titanium
Explanation:
Nitinol is an alloy of Nickel and Titanium. It posesses two properties such that,
- The shape memory effect
- Super elasticity
Shape memory is the ability of nitinol to undergo deformation at one temperature, stay in its deformed shape when the external force is removed.
Superelasticity is the ability for the metal to undergo large deformations and immediately return to its undeformed shape upon removal of the external load.
Hence, the correct option is (b) "Nickel and Titanium".
Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
The answer to this question would be: alkaline earth metal
Alkali earth metal is the second column group of the periodic table. In this group, the element has 2 extra electrons in their outer cells. That is why most of this metal has 2+ charge.
Their neighbor is the alkali metal which was the first column of the periodic table. The name is similar so don't confused and mix them each other.
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
