Answer:
C.) That all matter was composed of earth, fire, water and air
Explanation:
Just took the test
<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
Answer:
liquid
a semi permeable membrane
oxygen
Explanation:
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Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
