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elixir [45]
3 years ago
12

If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?

Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M
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Maslowich

Answer:

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7 0
3 years ago
How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ
Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

3 0
3 years ago
PLEASE HELP ME THIS IS DUE RN I HAVE NO IDEA I WILL MARK BRAINLIEST I JUST NEED HELP
djverab [1.8K]

Answer:

3.45 × 10^5

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6 0
3 years ago
The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st
Andrei [34K]
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2  so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄      ⇄     H⁺   + HC₂O₄⁻
0.0356 M            0          0
0.0356 - x            x          x
Ka1 = \frac{[H^+][HC2O4^-]}{[H2C2O4]} = x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
5 0
3 years ago
The densities of crystalline co2 and nh3 at 160 k are 1.56 and 0.84 g/cm3, respectively. calculate their molar volumes.
Neporo4naja [7]
The molar volume, symbol Vm<span>, is the </span>volume occupied by one mole of a substance at a given temperature and pressure. <span>It is equal to the </span>molar<span> mass divided by the mass density. Therefore, we calculate as follows:

Vm(CO2) = 44.01 / 1.56 = 28.21 cm^3 / mol
</span>Vm(NH3) = 17.03 / 0.84 = 20.27 cm^3 / mol
7 0
3 years ago
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