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3 years ago

Which of the following is an oxidation-reduction reaction?

Chemistry

1 answer:

garik1379 [7]3 years ago

6 0

Answer:

D) Ba(s) + 2 HCl(aq) → BaCl2(aq) + H2(g)

Explanation:

In a redox reaction, One atom is gaining electrons (Reduction) whereas other is loosing electrons (Oxidation). In the reactions:

A) HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l)

As sodium is always +1 and Br always -1, this is not a redox reaction.

B) NaI(aq) + AgNO3(aq) → AgI(s) + NaNO3(aq)

Sodium is always +1 and nitrate ion is -1. This is not a redox reaction

C) Pb(C2H3O2)2(aq) + 2 LiBr(aq) → PbBr2(s) + 2 LiC2H3O2(aq)

Acetate ion, C2H3O2 is -1 as Br is -1, lithium is +1 and Pb +2, This is not a redox reactoin

Barium is 0 as Ba(s) but +1 as BaCl2. Is oxidizing

Hydrogen is +1 as HCl but 0 as H2. Is reducting.

This reaction is an oxidation-reduction reaction

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If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

6 0

3 years ago

A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of

Goryan [66]

Answer:

C₂ H₄ O

Explanation:

1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)

atomic mass of C: 12.0107 g/mol

molar mass of CO₂: 44.01 g/mol

Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂

Solve for x:

x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

2) Mass of hydrogen (H) in 0.930 g of water (H₂O)

atomic mass of H: 1.00784 g/mol

molar mass of H₂O: 18.01528 g/mol

proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O

Solve for x:

x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

3) Mass of oxygen (O) in 1.0857 g of pure sample

Mass of O = mass of pure sample - mass of C - mass of H

Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

4) Mole calculations

Divide the mass in grams of each element by its atomic mass:

C: 0.59767 g / 12.0107 g/mol = 0.04976 mol

H: 0.10406 g / 1.00784 g/mol = 0.10325 mol

O: 0.3840 g / 15.999 g/mol = 0.02400 mol

5) Divide every amount by the smallest value (to find the mole ratios)

C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2

H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4

O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

C₂ H₄ O ← answer

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