Answer:
Half-life = 3 minutes
Explanation:
Using the radioactive decay equation we can solve for reaction constant, k. And by using:
K = ln2 / Half-life
We can find half-life of polonium-218
Radioactive decay:
Ln[A] = -kt + ln [A]₀
Where:
[A] could be taken as mass of polonium after t time: 1.0mg
k is Reaction constant, our incognite
t are 12 min
[A]₀ initial amount of polonium-218: 16mg
Ln[A] = -kt + ln [A]₀
Ln[1.0mg] = -k*12min + ln [16mg]
-2.7726 = - k*12min
k = 0.231min⁻¹
Half-life = ln 2 / 0.231min⁻¹
<h3>Half-life = 3 minutes</h3>
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Answer:
metal ball that moves at high speed
Explanation:
Answer: from what i can tell it looks like a heterogeneous mixture
Explanation:
heterogeneous is multiple different components, homogenous includes only the same component
Answer: Mass of 
required to form 930 kg of iron is 1328 kg
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
For iron:
Given mass of iron = 930 kg = 930000 g (1kg=1000g)
Molar mass of iron = 56 g/mol
Putting values in equation 1, we get:
The chemical equation for the production of iron follows:
By Stoichiometry of the reaction:
2 moles of iron are produced by = 1 mole of
So, 16607 moles of iron will be produced by = 
of
Now, calculating the mass of from equation 1, we get:
Mass of = 
Thus mass of required to form 930 kg of iron is 1328 kg
