K = 0.058 yr (constant)
Fraction of the pesticide = 36 years/12 = 3 half-lives , so around 1/8 left
when 99.9 % decomposed, 0.1 % is left , so the fraction is 0.001. To reach this point, it will take around +/- 120 years
Answer:
The answer is
<h2>877 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
density = 19.3 g/cm³
volume = 44.9 cm³
The mass of the gold bar
mass = 19.3 × 44.9 = 866.57
We have the final answer as
<h3>877 g</h3>
Hope this helps you
Usually, a model is a depiction of a certain entity, never the real thing. In times
past, many, many models look great on paper but are way off!
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
#SPJ4
Answer:
First one is: ammonia
Second one is: calcium hydroxide
Explanation: