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DedPeter [7]
3 years ago
15

In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al

l that apply)
1. HCH3CO2(aq) + NH3(aq) → CH3CO−2(aq) + NH+4(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base

2. BH3(aq) + NH3(aq) → BH3NH3(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base

3. HNO2(aq) + C2H5NH2(aq) → NO−2(aq) + C2H5NH+3(aq)
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
Chemistry
1 answer:
tekilochka [14]3 years ago
7 0

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)

here underlined is HCH_3CO_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

Here underlined is NH_3

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

Here underlined is C_2H_5NH_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1.  HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)

As HCH_3CO_2(aq) is donating a proton , it acts as a bronsted acid.

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

As NH_3 contains a lone pair of electron on nitrogen , it can easily donate electrons to BH_3 and act as lewi base.

3.  HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

As C_2H_5NH_2(aq) is accepting a proton , it acts as a bronsted base.

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How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
IgorC [24]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Converting mass of Al_{2}(S_{2}O_{3})_{3}to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Converting mol Al_{2}(S_{2}O_{3})_{3}to mol S:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Converting mol S to atoms of S using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
3 years ago
Solve the simultaneous equation<br> -4x+3y=1<br> 6x-y=2
Maslowich

Answer:

x = 1/2 y = 1

5 0
1 year ago
Read 2 more answers
Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

7 0
4 years ago
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4 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
ludmilkaskok [199]

Answer:

1.20 V

Explanation:

Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Given,

Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)

E^0_{[Pb^{2+}/Pb]}= -0.13\ V

Br_2(l) + 2 e^-\rightarrow 2 Br(aq)

E^0_{[Br_2/Br^{-}]}=+1.07\ V

E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}

E^0=+1.07- (-0.13)\ V=1.20\ V

6 0
4 years ago
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