Monosaccharides are the simplest carbohydrates. Although glucose and fructose have the same molecular formula they have different structures. They cannot be further hydrolyzed to simple sugars. Disaccharides contains two monosaccharides. For example, lactose and sucrose. Polysaccharides on the other hand contains a large number of saccharides. An example is starch, glycogen and dextrans. Amino acids contains an amino acid, carboxyl group and an R-group. Whatever the diagram you have, you just look at the structures contained.
Answer:
It is called a salt bridge
To calculate percent errorsubtract the accepted value from the experimental value.Take the absolute value of step 1.Divide that answer by the accepted value.Multiply that answer by 100
so% error = [|(10.085 g/ 10 ml) - 0.9975| / 0.9975] x 100
% error = 1.1 %
Answer:
2.79 mg iron-59 is left in the sample after 267 days.
Explanation:
Given that:
Half life = 44.5 days
Where, k is rate constant
So,
The rate constant, k = 0.0156 days⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 180 mg
Time = 267 days
So,
![[A_t]=180\ mg\times e^{-0.0156\times 267}](https://tex.z-dn.net/?f=%5BA_t%5D%3D180%5C%20mg%5Ctimes%20e%5E%7B-0.0156%5Ctimes%20267%7D)
![[A_t]=180\times e^{-0.0156\times 267}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D180%5Ctimes%20e%5E%7B-0.0156%5Ctimes%20267%7D%5C%20mg)
![[A_t]=2.79\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D2.79%5C%20mg)
<u>2.79 mg iron-59 is left in the sample after 267 days.</u>
The volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm is 7.81L.
<h3>HOW TO CALCULATE VOLUME?</h3>
The volume of a gas can be calculated using the following formula:
PV = nRT
Where;
- P = pressure (atm)
- V = volume (L)
- R = gas law constant (0.0821 Latm/molK)
- n = number of moles
- T = temperature (K)
V = 0.323 × 0.0821 × 265 ÷ 0.900
V = 7.027 ÷ 0.9
V = 7.81L
Therefore, the volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm is 7.81L.
Learn more about volume at: brainly.com/question/1578538
