Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.
So first, we take our measured value, .299 cm, minus our known value, .225 cm.
.299 cm - .225 cm=.004 cm
Next, we divide that by our known value

Finally, multiply your answer by 100
.0177777778 x 100= 1.77777778 %
Round to three significant figures, and you're done.
=1.78 % error
Answer:
Option A. KCl (aq)
Option D. Mg(OH)₂(s
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂(aq) + KOH(aq) —>
In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
MgCl₂(aq) + KOH(aq) —>
Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)
MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)
Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)
Thus, option A and D gives the correct answer to the question.
The overall reaction is given by:

The fast step reaction is given as:

The slow step reaction is given as:
(slow step
)
Now, the expression for the rate of reaction of fast reaction is:
![r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=r_%7B1%7D%3Dk_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D-k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
The expression for the rate of reaction of slow reaction is:
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as
takes place in this reaction.
The expression of rate of formation is:

=
(1)
Now, consider that the fast step is always is in equilibrium. Therefore, 
![k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=k_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D%3D%20k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
![[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D%20%3D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D)
Substitute the value of
in equation (1), we get:
![\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28NOBr%29%7D%7Bdt%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
=![k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D%5BNO%5D)
= ![\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7B1%7Dk_%7B2%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D)
Thus, rate law of formation of
in terms of reactants is given by
.
Answer:
Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.
Answer:
moles of glucose
<u>2.3166 moles of glucose</u>
<u></u>
Explanation:
The balance reaction for the formation of glucose is :

here , CO2 = carbon dioxide
H2O = water
C6H12O6 = glucose
O2 = Oxygen
According to this equation :
6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2
We are asked to calculate the mole of Glucose from carbon dioxide.
So,
6 mole of CO2 produce = 1 mole of C6H12O6
1 mole of CO2 will produce =
moles of glucose
13.9 moles of CO2 will produce :

=2.3166 moles of glucose
Note : first , Always calculate for one mole (By dividing)
. After this , multiply the answer with the moles given.
Always write the substance whose amount is asked(glucose) to the right hand side