Always carry the microscope with________________-one on the ____________ and one underneath the___________

A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The net ionic equation for the neutralization reaction that oc

cluponka [151]

Answer:

The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .

Explanation:

HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.

IUPAC name of this compound is Ethanoic acid.

Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.

Given that, equivalence point was reached when 20.0mL of NaOH is added.

let the normality of acetic acid is N₁ and that of NaOH is N₂.

volume of acetic acid is V₁ and that of NaOH is V₂.

Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.

⇒ N₁ × V₁ = N₂ × 20.

after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).

= N₂ × 15.

after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).

= N₂ × 19.

⇒ V° : Vˣ = 15 : 19 .

⇒

7 0

3 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

Maru [420]

Explanation:

The given data is as follows.

$V_{1}$ = 50 ml, $T_{1}$ = 345 K

$T_{2}$ = 298 K, $T_{f}$ = 317 K,

$V_{2}$ = 50 ml

Now, we will calculate the heat energy as follows.

$Q_{hot} = m \times C_{p} \times (T_{1} - T_{f})$

= $50 g \times 4.184 \times (345 - 317)$

= 5857.6 J

$Q_{cold} = m \times C_{p} \times (T_{f} - T_{2})$

= $50 g \times 4.184 \times (317 - 298)$

= 3974.8 J

As, $Q_{hot} = -Q_{cold}$ so there will be loss of heat. And, some heat will go to the calorimeter.

Hence, $Q_{hot} = Q_{cold} + Q_{cal}$

$5857.6 = 3974.8 + Q_{cal}$

$Q_{cal}$ = 1882.8 J

We assume that the temperature of (calorimeter + water) is 298 K. Hence,

dT = $T_{f} - T_{2}$

= (317 - 298) K

= 19 K

Hence, we will calculate the specific heat as follows.

C = $\frac{Q}{dT}$

= $\frac{1882.8 J}{19}$

= 99.1 J/K

Thus, we can conclude that the value of $C_{cal}$ for the calorimeter is 99.1 J/K.

3 0

4 years ago

What is the ground-state electron configuration of a neutral atom of manganese?.

Alchen [17]

Explanation:

elctronic configuration of manganese

Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵

ground state

Mn=Ar3d⁵4s²

note that Ar is argon

3 0

3 years ago

The mass of an iron(II) sulfate crystal is 8.36 g.How many moles of FeSO4 are in the crystal?

Yuki888 [10]

The number of moles present in the FeSO4 are 0.055 mol.

Explanation:

The mass of a substance containing the same number atoms in 12.0 g of 12C is known as mole. One mole of any substance is equal to 6.023 x 10^23. The moles of a substance can be determined by using the formula,

Number of moles = mass in grams / molecular mass

Given,

mass = 8.36 g,

molecular mass of FeSO4 = 151.908 g / mol

number of moles = 8.36 / 151.908

= 0.055 mol.

5 0

4 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$