<span>The following is the role of NaCl and EDTA in DNA isolation, hope it helps:
NaCl provides Na+ ions that will block negative charge from phosphates on DNA.
Negatively charged phosphates on DNA cause molecules to repel each other. The Na+ ions will form an ionic bond with the negatively charged phosphates on the DNA, neutralizing the negative charges and allowing the DNA molecules to come together.</span>
Answer:
<h3>The answer is 7.50 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of iron = 150 g
volume = 20 mL
We have

We have the final answer as
<h3>7.50 g/mL</h3>
Hope this helps you
<span> the first ionization </span>energy<span> of an element is the </span>energy<span> needed to</span>remove<span> the outermost, or highest </span>energy<span>, </span>electron<span> from a neutral </span>atom<span> in the gas phase.</span>
Answer:
Double replacement
Precipitation reaction
Explanation:
You have the reaction:
REACTANTS PRODUCTS
BaCl₂ (aq) + Na₂SO₄ (aq) ⇒ 2NaCl (aq) + BaSO₄(s)
The general form of a double replacement reaction is the following:
AB + CD ⇒ CD + AB
The reactants basically, exchanged partners. In the case of your problem, Barium(Ba) and Sodium(Na) switched places. So this makes it a double-replacement reaction.
Now how do I know it is a precipitation reaction. A precipitation reaction occurs when two solutions combine and salt is formed. Salt is solid, so how do I know that's what occured? Look at your equation again:
BaCl₂ (aq) + Na₂SO₄ (aq) ⇒ 2NaCl (aq) + BaSO₄(s)
aq means aqueous (liquid)
s means solid
If you look at the product formed in the reaction, from two solutions, it formed a solid. So this is your clue as to why it is a precipitation reaction.
<span>The </span>standard enthalpy of formation<span> <span>is defined as the change in </span></span>enthalpy<span> <span>when one mole of a substance in the </span></span>standard<span> <span>state (1 atm of pressure and temperature of 298.15
K) is </span></span>formed<span> <span>from its
pure elements under the same conditions.</span></span>