The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of silver = 16.2g / 107.8g/mol = 0.15mol
From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.
Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g
Hence required mass of silver nitrate is 25.48g.
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The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.
<h3>What is equilibrium?</h3>
Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
A very high value of K indicates that at equilibrium most of the reactants are converted into products.
The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.
When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.
This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.
Hence, option A is correct.
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The following observation would be included in a hypothesis: a connection between type of material and thermal energy transfer is developed (option C).
<h3>What is a hypothesis?</h3>
Hypothesis is a tentative conjecture explaining an observation, phenomenon or scientific problem that can be tested by further observation, investigation and/or experimentation.
The hypothesis is regarded as an educated guess because it predicts the possible outcome of an experiment.
The hypothesis is a statement that relates the independent variable of the experiment with the measured or dependent variable.
Therefore, the following observation would be included in a hypothesis: a connection between type of material and thermal energy transfer is developed.
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Answer:
1 . 
2. 
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which 
behaves as cation and 
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation 
is smaller than . Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation 
is greater than that of and also the size of the cation is smaller than . Thus, this corresponds to higher lattice energy.
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
