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3 years ago

Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to neutralize it?

1 answer:

3 0

In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.

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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Scilla [17]

Answer:

The essence including its given problem is outlined in the following segment on the context..

Explanation:

The given values are:

Moles of CO₂,

x = 0.01962

Moles of water,

$\frac{y}{2} =0.01961$

$y=2\times 0.01961$

$=0.03922$

Compound's mass,

= 0.4647 g

Let the compound's formula will be:

$C_{x}H_{y}O_{z}$

Combustion's general equation will be:

⇒ $C_{x}H_{y}O_{z}+x+(\frac{y}{4}-\frac{z}{2}) O_{2}=xCO_{2}+\frac{y}{2H_{2}O}$

On putting the estimated values, we get

⇒ $12\times x=1\times y+16\times z=0.4647$

⇒ $12\times 0.01962+1\times 0.03922+16\times z=0.4647$

⇒ $0.27466+16z=0.4647$

⇒ $z=0.01187$

Now,

x : y : z = $0.01962:0.03922:0.01187$

= $\frac{0.01962}{0.0118}:\frac{0.03922}{0.0188}:\frac{0.0188}{0.0188}$

= $1.6:3.3:1.0$

= $3:6:2$

So that the empirical formula seems to be "C₃H₆O₂".

8 0

3 years ago

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Ivanshal [37]

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

5 0

3 years ago

Help mark brainliest !! Both dehydration and water addition to alkenes use h2so4 as a catalyst.Which is the difference between t

sasho [114]

Can u take a picture of it

7 0

3 years ago

Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 per

Eduardwww [97]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure water = $0^oC$

$T_f$ = temperature of solution = $-0.89^oC$

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}$

$\text{Molar mass of substance(solute)}=157.3g/mol$

Therefore, the molecular weight of a substance is 157.3 g/mol

7 0

3 years ago

The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r

Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

5 0

3 years ago