Answer:
The correct answer is "1.0100".
Explanation:
Let the volume of mixture be 100 ml.
then,
The volume of DMSO will be 10 mL as well as that of water will be 90 mL.
DMSO will be:
= 
= 
The total mass of mixture will be:
= 
= 
Density of mixture will be:
= 
= 
= 
hence,
Specific gravity of mixture will be:
= 
= 
= 
Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:
There are 59.2 grams of water in this solution.
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
Answer:
Each specific property of motif and domain is explained.
Explanation:
Domain;
- May retain a 3D structure when separated from rest of the protein.
- Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
- Stable globular units like pyruvate kinase
- May be distinct functional units in a protein
Motif;
- Repetetive supersecondary structure because they contain cluster of secondary structure.
- Beta Alpha Beta unit is an example of motif
- Clusters of secondary structure
Both Motif and Domain;
- Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
- Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.
