Answer:
50 g of K₂CO₃ are needed
Explanation:
How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?
We analyse data:
500 g is the mass of the solution we want
10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution
Therefore we can solve this, by a rule of three:
In 100 g of solution we have 10 g of K₂CO₃
In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃
Answer:
1.196 M NaOH
Explanation:
Molarity = moles/Volume (L)
moles NaOH = mass NaOH/MM NaOH = 12/40.01 = 0.299 moles NaOH
Volume solution = 250 mL = 0.250L
M = 0.299/0.250=1.196 M NaOH
I think it's 6?...........
Answer:
The correct option is: non-covalent interactions
Explanation:
Enzymes are the macromolecules that are responsible for catalyzing biochemical reactions in the cell and thus they are known as the<u> biological catalysts.</u> They have<u> high specificity and selectivity.</u>
In an enzyme-catalyzed reaction, the substrate molecules bind to the active sites present on the surface of the enzyme, resulting in the formation of a enzyme-substrate complex, which is stabilized by weak non-covalent interactions such as Van der Waals forces, hydrophobic interactions, electrostatic forces and hydrogen bonding.
Answer:
The molarity of the solution is: 2 M
Explanation:
We calculate the weight of 1 mol of NaOH:
Weight 1 mol NaOH= Weight Na + Weight 0 + Weight H= 23 g+16g+ 1 g= 40g/mol
The molarity is the quantity of moles in 1000ml (1 liter) of solution:
0,5liter-----40 grams NaOH
1 liter----x=(1 liter x 40 grams NOH)/0,5 liter = 80 grams NaOH
40 grams----1 mol NaOH
80 grams----X=(80 grams x 1 mol NaOH)/40 grams= 2 mol NaOH--> 2M
