Answer:
A mixture whose components are soluble in each other. ... a solution that has water as its solvent; most have an ionic substance as the solute, may contain a liquid ... The suspensions of particles larger than individual ions or molecules, but the ... This effect is used to determine whether a mixture is a true solution or a colloid.
process of solute particles being surrounded by water molecules arranged in a ... solution. homogeneous mixture consisting of a solute dissolved into a solvent. ... apart from the crystal, the individual ions are then surrounded by solvent particles in a ... are intermediate in size between those of a solution and a suspension.
A suspension is a heterogeneous mixture in which some of the particles ... The particles in a suspension are far larger than those of a solution, so gravity is … ... Particle size: 0.01-1nm; atoms, ions or molecules, Particle size: ... solutions because the individual dispersed particles of a colloid cannot be seen.
Explanation:
Answer:
0.25L
Explanation:
Using the dilution formula
C1V1=C2V2
C1=6M
V1?
C2=0.75M
V2=2.0L
V1= C2V2/C1
V1=0.75*2.0/6
V1=0.25L
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Higher probability of loss. Chorionic villus sampling (CVS) and Amniocentesis (AC). The prenatal diagnosis technique can be done earlier in fetal development CVS (first trimester --> 10-13 weeks). AC (second trimester --> 16-20 weeks)
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Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles =
Number of moles =
= 0.003mole