Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
Pls mark it as branliest answere thanks
Find it on google i’m pretty sure i saw it somewhere so sorry this doesn’t help
Answer:
K = 137.55 atm/M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = (K)(C)</em>
where P is the partial pressure of the gaseous solute above the solution (P = 1.0 atm).
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).
∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.
Since its volume you do what it says 55 cm x 100 cm x 80cm and do 100 x 50 which is 5,000. then 5,000 x 80 which is 40,000. sorry i cant show work on here.
Answer:
Once an enzymatic reaction is completed, the enzyme releases substrates.
Explanation:
The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).
