Answer:
16.791 grams
Explanation:
The density formula is:

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.


The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

Substitute the values into the formula.


Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.


The mass of the gold nugget is 16.791 grams.
Answer:
A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula
. It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water
Explanation:
Given data includes:
Tris= 10mM
pH = 8.0
NaCl = 150 mM
Imidazole = 300 mM
In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.
Stock Concentration Volume to be Final Concentration
added
1 M Tris 2.5 mL 10 mM
5 M NaCl 7.5 mL 150 mM
1 M Imidazole 75 mL 300 mM
. is the formula that is used to determine the corresponding volume that is added for each stock concentration
The stock concentration of Tris ( 1 M ) is as follows:
.

The stock concentration of NaCl (5 M ) is as follows:
.

The stock concentration of Imidazole (1 M ) is as follows:
.

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.
We can store the copper sulphate solution in alumiun container, if cover on alumiun is present.
<h3>Can you store cuso4 in an aluminum container?</h3>
Aluminium is more reactive than copper so the Aluminium will displace copper sulphate from its solution by reacting with it but if there is cover on the aluminium then the alumium can't react with copper.
So we can store the copper sulphate solution in alumiun container.
Learn more about container here: brainly.com/question/11459708
There are a total of 4 elements
The balanced chemical reaction is:
<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.
2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3
4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3
63.8g N2H4 <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>