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3 years ago

Which statement explains why 10.0 mL of a 0.50 M H2SO4(aq) solution exactly neutralizes 5.0 mL of a 2.0 M NaOH(aq) solution?

Chemistry

1 answer:

Mandarinka [93]3 years ago

4 0

Answer:

(1) The moles of H+(aq) equal the moles of OH-(aq).

Explanation:

Statements are:

(1) The moles of H+(aq) equal the moles of OH-(aq).

(2) The moles of H2SO4(aq) equal the moles of NaOH(aq).

(3) The moles of H2SO(aq) are greater than the moles of NaOH(aq).

(4) The moles of H+(aq) are greater than the moles of OH-(aq).

Neutralization reaction of H₂SO₄ with NaOH is:

H₂SO₄ + 2 NaOH → 2H₂O + Na₂SO₄

Where 1 mole of sulfuric acid (2 moles H⁺), reacts with 2 moles of NaOH (2moles of OH⁻)

As you cans see, the reaction occurs when moles of H⁺ are equal to moles of OH⁻

In the problem, moles of H⁺ are:

0.010L ₓ (0.50moles H₂SO₄ / 1L) ₓ (2moles H⁺ / mole H₂SO₄) =

0.01 moles of H⁺

Moles of OH⁻ are:

0.005L ₓ (2.0moles NaOH / 1L) ₓ (1mole OH⁻ / mole NaOH) =

0.01 moles of OH⁻

Thus, right statement is:

(1) The moles of H⁺(aq) equal the moles of OH⁻(aq). 

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How long have orca whales been on the endangered species list?

kenny6666 [7]

1988 I think so or 1981

6 0

4 years ago

8.0 g of cream at 17.0 °C are added to an insulated cup containing 150.0 g of coffee at 73.2 °C. Calculate the equilibrium tempe

Anna11 [10]

Answer:

70.35°C

Explanation:

From the question,

Assuming: The specific heat capacity of cream and coffee is thesame as that of water = 4.2 J/g.°C

Applying

Heat lost by the coffee = heat gained by the cream

cm(t₁-t₃) = cm'(t₃-t₂)................ Equation 1

Where c = specific heat capcity of both cream and coffee, m = mass of coffee, t₁ = intial temperature of coffee, m' = mass of cream, t₂ = initial temperature of cream, t₃ = equilibrium temperature

From the question,

Given: m = 150 g, t₁ = 73.2°C, m' = 8 g, t₂ = 17°C

Constant: c = 4,2 J/g.°C

Substitute these values into equation 1

4.2×150(73.2-t₃) = 4.2×8(t₃-17)

630(73.2-t₃) = 33.6(t₃-17)

Solve for t₃

46116-630t₃ = 33.6t₃-571.2

630t₃+33.6t₃ = 571.2+46116

663.6t₃ = 46687.2

t₃ = 46687.2/663.6

t₃ = 70.35°C

6 0

3 years ago

The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions

musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,

x as 0.115 and y as 0.165

Now the mass of AgCl will be,

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,

278.1 × 0.165 = 45.88 grams.

3 0

3 years ago

Convert 18.63 mol Nicly to grams.

ddd [48]

Answer:

1093.458042 grams

Explanation:

1 mole is equal to 1 moles Nickel, or 58.6934 grams. So I think you can unsterstand how I did the math if you have any questions let me know.

6 0

3 years ago

Naming Compounds Test Review Practice Name the following ionic compounds: NH4Cl _____________________________________ Fe(NO3)3 _

Mariulka [41]

Answer :

The name of $NH_4Cl$ is ammonium chloride.

The name of $Fe(NO_3)_3$ is iron(III) nitrate.

The name of $TiBr_3$ is titanium(III) bromide.

The name of $Cu_3P$ is copper(I) phosphide.

The name of $SnSe_2$ is tin(IV) selenide.

The name of $GaAs$ is gallium arsenide.

The name of $Pb(SO_4)_2$ is lead(IV) sulfate.

The name of $Be(HCO_3)_2$ is beryllium hydrogen carbonate.

The name of $Mn_2(SO_3)_3$ is manganese(III) sulfite.

The name of $Al(CN)_3$ is aluminium cyanide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) $NH_4Cl$

It is an ionic compound because it is formed by the combination of one polyatomic cation (ammonium ion) and anion (chloride ion). The charge on ammonium ion is (+1) and the the charge on chloride ion is (-1).

Thus, the name of $NH_4Cl$ is ammonium chloride.

(b) $Fe(NO_3)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (nitrate ion) and cation (iron ion). The charge on iron ion is (+3) and the the charge on nitrate ion is (-1).

Thus, the name of $Fe(NO_3)_3$ is iron(III) nitrate.

(c) $TiBr_3$

It is an ionic compound because it is formed by the combination of titanium cation and bromide anion. The charge on titanium is (+3) and the the charge on bromide ion is (-1).

Thus, the name of $TiBr_3$ is titanium(III) bromide.

(d) $Cu_3P$

It is an ionic compound because it is formed by the combination of copper cation and phosphorous anion. The charge on copper is (+1) and the the charge on phosphorous ion is (-3).

Thus, the name of $Cu_3P$ is copper(I) phosphide.

(e) $SnSe_2$

It is an ionic compound because it is formed by the combination of tin cation and selenium anion. The charge on tin is (+4) and the the charge on selenium ion is (-2).

Thus, the name of $SnSe_2$ is tin(IV) selenide.

(f) $GaAs$

It is an ionic compound because it is formed by the combination of gallium cation and arsenic anion. The charge on gallium is (+3) and the the charge on arsenic ion is (-3).

Thus, the name of $GaAs$ is gallium arsenide.

(g) $Pb(SO_4)_2$

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfate ion) and cation (lead ion). The charge on lead ion is (+4) and the the charge on sulfate ion is (-2).

Thus, the name of $Pb(SO_4)_2$ is lead(IV) sulfate.

(h) $Be(HCO_3)_2$

It is an ionic compound because it is formed by the combination of one polyatomic anion (hydrogen carbonate ion) and cation (beryllium ion). The charge on beryllium ion is (+2) and the the charge on hydrogen carbonate ion is (-1).

Thus, the name of $Be(HCO_3)_2$ is beryllium hydrogen carbonate.

(i) $Mn_2(SO_3)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfite ion) and cation (manganese ion). The charge on manganese ion is (+3) and the the charge on sulfite ion is (-2).

Thus, the name of $Mn_2(SO_3)_3$ is manganese(III) sulfite.

(j) $Al(CN)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (cyanide ion) and cation (aluminium ion). The charge on aluminium ion is (+3) and the the charge on cyanide ion is (-1).

Thus, the name of $Al(CN)_3$ is aluminium cyanide.

3 0

3 years ago