Answer:
The two enzymes that are used during DNA replication is helicase and polymerase. Each enzyme has distinct role that made it a vital part of the replication. In the course of DNA replication, the initial stage is to unzip the double helix structure of the DNA molecule by the enzyme helicase, which breaks the hydrogen bonds that holds the complementary bases of DNA.
Moreover, the DNA polymerase has the ability to exactly copy a DNA template. This enzyme binds to the leading strand and then adding a new complementary nucleotide bases to the DNA strand. In addition, it catalyzes the joining of deoxyribonucleoside 5′-triphosphates (dNTPs) to form the increasing DNA chain.
Explanation:
<span>Take the inversion of density: 1mL/13.6 g and multiply it by the conversion factor 453.6 g/ 1 lb and the given 5.00 lb. Units for mass (grams) and units for weight (lbs) cancel leaving only units of volume. I believe it should be 167 mL or 0.167 L</span>
In flame tests salts that are dissolved in water are evaporated using a hot flame. In the flame the metal atoms become excited and produce their characteristic spectrum of light. Hope this answers the question. Have a nice day. Feel free to ask more questions.
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.