The half-life of Th-232 is 1.405 × 10¹⁰ years
Time elapsed = 2.8 x 10⁹ years
Equation of radioactive decay:
A = A₀ = (1/2)^ t/t₁/₂
Thus, the percentage of thorium-232 in the rock that was dated at 2.8 billions year = 87.1%
They have high melting and boiling points and are hard and brittle
Answer:
0.1593 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
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
