We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
Answer:
C. It decreases by a factor of 4
Explanation:
F1 = kq1*q2/r²
F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4
It is a Compound. have a nice day
Answer:
See below
Explanation:
propane mole weight = 44 gm / mole
100 gm / 44 gm / mole = 2.27 moles
From the equation, 5 times as many moles of OXYGEN (O2)are required
= 11.36 moles of oxygen
at <u>STP</u> this is 254.55 liters of O2 (because 22.4 L = one mole) and
Using oxygen as 21 percent of air means that
.21 x = 254.55 = x = <u>1212.12 liters of air required </u>
Answer:
B-Sucrose molecules are too large to conduct electricity in once dissolved in water.
D-Salts, like NaCL, have ionic bonds and are considered to be electrolytes:when dissolved in water, salts dissociate and form ions.
Explanation:
