Sucrose, a sweet, white crystalline substance, C12 H22 O11, OBTAINED CHIEFLY FROM THE JUICE OF THE SUGAR CANE AND SUGAR BEET, BUT ALSO PRESENT IN SORGHUM, THE sugar maple, some palms, and various other plants, and having extensive nutritional, pharmaceutical, and industrial uses; any of the class of carbohydrates to which this substance belongs, as glucose, levulose, and lactose.
The question is incomplete. The complete question is :
A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)
Solutions :
If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.
Let's take :


The rate constant =
respectively.
The activation energy and the Arhenius factor is same.
So by the arhenius equation,
and 




Given,
J/mol
R = 8.314 J/mol/K





∴ 
So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.
Answer:
M = 20.5 g/mol
Explanation:
Given data:
Volume of gas = 1.20 L
Mass of gas = 1.10 g
Temperature and pressure = standard
Solution:
First of all we will calculate the density.
Formula:
d = mass/ volume
d = 1.10 g/ 1.20 L
d = 0.92 g/L
Now we will calculate the molar mass.
d = PM/RT
0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K
M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm
M = 20.5 g/mol
8 moles of H 2O are produced.
First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.
<h3>How many moles of H2 are required to make one mole of NH3?</h3>
Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.
<h3>
How many moles of h2 are needed to produce a solution?</h3>
An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).
Learn more about H2O:
brainly.com/question/2193704
#SPJ4