Bromine water can be used to distinguish between these two molecules?? true or false

Question 3 When the population size decreases over time, the total number of births the total number of deaths. O less than O eq

Neporo4naja [7]

The population decreases when the birth rate is less than the death rate.

4 0

3 years ago

Need help on both #19 and #20 please

ziro4ka [17]

This is the answer to question 19. If it’s not clear, send me a message.

5 0

4 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

Nezavi [6.7K]

Answer:

The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

mass = 28 mg = 0.028 g
molar mass = 90.03 g/mol
Convert mass in grams to number of moles, n:

n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol

4) Titration

Volume of base: 28.4 mL = 0.0248 liter
Concentration of base: x (unknwon)

Number of moles of acid: 2.52 mol (calculated above)

Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

5) Molarity of NaOH solution

M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

5 0

3 years ago

Read 2 more answers

Water that has low saline content is called ________

ale4655 [162]

It is called fresh water.

6 0

4 years ago

The student adds 6 m nh3 to be cuso4 solution until the cu2+ ion is essentially all converted to cu(nh3)42+ ion. the voltage of

Gnoma [55]

Here we have to get the residual concentration of Cu²⁺ in the reaction.

The residual concentration of [Cu²⁺] is 2.137×10⁸ m

The reaction between ammonia (NH₃) and CuSO₄ in ionic form is - Cu²⁺ + 4 NH₃ = [Cu(NH₃)₄]²⁺.

Thus 1 mole of Cu²⁺ will react with 4 moles of NH₃.

Now 6m NH₃ or 6 moles of ammonia will react with $\frac{6}{4}$ = 1.5 moles of Cu²⁺.

Thus the Nernst equation of the reaction is $E_{Cell}$ = E° + $\frac{0.059}{n}$ log $\frac{Cu^{2+} }{Cu}$ .

Here the number of moles is 3, as 6 moles of ammonia will react with 1.5 moles of cu²⁺. and the standard reduction potential (E°) of Cu²⁺/Cu system is (+) 0.674V and the cell potential is 0.92V

On plugging the values in the equation, we get,

0.92 = 0.674 + $\frac{0.059}{2}$ log[Cu²⁺]

Or, 0.029 log [Cu²⁺]= 0.246

Or, log [Cu²⁺] = 8.33

Or, [Cu²⁺] = 2.137×10⁸

3 0

3 years ago

Bromine water can be used to distinguish between these two molecules?? true or false​

Bromine water can be used to distinguish between these two molecules?? true or false