Answer:- 
Explanations:- It is given that the charge for A is +2 and the charge for B is -3. The over all compound is neutral means the over all charge is zero. For making the over all charge zero, we need 3 positive ions and 2 negative ions. This makes a +6 charge for A and -6 charge for B and the over all charge is zero.
Also, if we think about the criss cross then charge of A becomes the subscript of B and the charge of B becomes the subscript of A.
So, the formula of the ionic compound is . In this compound the ratio of A to B is 3:2.
The balanced equation could be shown as:
Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:
c) Half life for first-order kinetics is computed by:
d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:
Best regards.
Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).
The reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.
However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.
This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.
The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.
Answer:
The final pressure is 0.725 atm.
Explanation:
Gay Lussac's Law establishes the relationship between pressure and temperature of a gas when the volume is constant. This law says that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, pressure and temperature are directly proportional quantities.
Mathematically, Gay-Lussac's law states that, when a gas undergoes a constant volume transformation, the quotient of the pressure exerted by the gas temperature remains constant:
When analyzing an initial state 1 and a final state 2, the following is satisfied:
In this case:
- P1= 0.81 atm
- T1= 33 C= 306 K
- P2= ?
- T2= 1 C= 274 K
Replacing:
Solving:
P2= 0.725 atm
<u><em>The final pressure is 0.725 atm.</em></u>
