Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
The answer for the following problem is mentioned below.
- <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
Explanation:
Given:
no of moles of the oxygen gas = 0.692
Also given:
2 HgO → 2 Hg + 
where,
HgO represents mercuric oxide
Hg represents mercury
represents oxygen
To calculate:
Molar mass of HgO:
Molar mass of HgO = 216 grams
molar mass of mercury (Hg) = 200 grams
molar mass of oxygen (O) =16 grams
HgO = 200 +16 = 216 grams
We know;
2×216 grams of HgO → 1 mole of oxygen molecule
? → 0.692 moles of oxygen molecule
= 
= 298.944 grams of HgO
<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
<u />
Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
Explanation:
Scientist use trees a whole lot to look at climate of the past by examining tree rings.
These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.
Tree rings can be used to decipher the age of a tree.
- These three rings can be used to interpret climatic patterns.
- During a wet climate, the tree rings are more robust and bigger.
- In a dry climate, the rings are thinner.
- These alternating patterns can be used to decipher the climatic signatures in a tree.
- Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.
learn more:
Climate change brainly.com/question/7824762
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<h3>
Answer:</h3>
C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)
<h3>
Explanation:</h3>
The balanced chemical equation for the combustion of the hydrocarbon in question is;
C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)
- A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
- Reactant side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
- Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
- An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
- An equation is balanced so as to obey the law of conservation of mass.