Answer:
No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 =
or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
<span>Sugar is not soluble in Oil so it coagulated when Joyce added sugar to oil.
But sugar is soluble in water so when Joyce added water in the mixture of OIl and Sugar and then shaked it the sugar dissolved in the water.Leaving no traces behind.As Oil is hydrophobic it doesn't mix with water instead it floats on top.So by using the solubility of sugar in water Joyce separated sugar from oil</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.