Answer:
It is possible she could get one.
Explanation:
To solve this problem we need to convert 98.3 kilometers/hour to miles/hour.
In other words, we <u>convert km to mi</u>, to do so we multiply 98.3 km by a <em>conversion factor</em>, putting the unit we want to have in the numerator, and the unit we want to convert in the denominator:
- 98.3 km *
= 61 mi
Given that the little old lady is doing 61 miles/hour, she could get a speeding ticket.
Answer:
% purity of limestone = 96.53%
Explanation:
Question (4).
Weight of impure CaCO₃ = 25.9 g
Molecular weight of CaCO₃ = 40 + 12 + 3(16)
= 100 g per mole
We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters
From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of
CO₂.
∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g
∴ 1 liter of CO₂ will be produced by CaCO₃ = 
∴ 5.6 liters of CO₂ will be produced by CaCO₃ = 
= 25 g
Therefore, % purity of CaCO₃ = 
= 
= 96.53 %
Answer:
The pressures will remain at the same value.
Explanation:
A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.
While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.
Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).
The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.
Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.
From other sources, the given mass of the solute that is being dissolved here is 7.15 g Na2CO3 - 10H2O. We use this amount to convert it to moles of Na2CO3 by converting it to moles using the molar mass then relating the ratio of the unhydrated salt with the number of water molecules. And by the dissociation of the unhydrated salt in the solution, we can calculate the moles of Na+ ions that are present in the solution.
Na2CO3 = 2Na+ + CO3^2-
7.15 g Na2CO3 - 10H2O (1 mol / 402.9319 g) (1 mol Na2CO3 / 1 mol Na2CO3 - 10H2O) ( 1 mol Na2CO3 / 1 mol Na2CO3-10H2O ) ( 2 mol Na+ / 1 mol Na2CO3) = 0.04 mol Na+ ions present