Ser Bachiller 1 (13th)

Ser Bachiller 1 (13th)

1st - 12th Grade

10 Qs

quiz-placeholder

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Ser Bachiller 1 (13th)

Ser Bachiller 1 (13th)

Assessment

Quiz

Chemistry

1st - 12th Grade

Hard

Created by

Adrian Huaca

Used 3+ times

FREE Resource

10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Iguale por redox la ecuación


HNO3 + Zn → Zn (NO3)2 + NH4(NO3) + H2O

10 HNO3 + 4 Zn → 4 Zn (NO3)2 + NH4(NO3) + 3 H2O

10 HNO3 + 4 Zn → Zn (NO3)2 + 4 NH4(NO3) + H2O

20 HNO3 + 2 Zn → 2 Zn (NO3)2 + 8 NH4(NO3) + 2 H2O

22 HNO3 + 8 Zn → 8 Zn (NO3)2 + 2 NH4(NO3) + 7 H2O

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Con base en la reacción, determine el agente reductor


FeCl2 + H2O2 + HCl → FeCl3 + H2O

Oxígeno

Hierro

Cloro

Hidrogeno

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Determine los números de oxidación que correspondan al elemento nitrógeno en la ecuación


Fe + HNO3 → Fe (NO3)3 + NH4NO3 + H2O

N = 5+, 3+

N = 3+, 5+

N = 5-, 3+

N = 5+, 3-

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Seleccione las ecuaciones químicas balanceadas por simple inspección


1. 4 HClO3 + Pb (OH)4 → Pb (ClO3)4 + 4 H2O

2. H2S + K (OH) → K2S + H2O

3. 3 H2CO3 + Bi (OH)3 → Bi2(CO3)3 + 6 H2O

4. H2SO4 + 2 Li(OH) → Li2(SO4) + 2 H2O

5. 2 HCl + Zn → ZnCl2 + H2

1, 2, 4

1, 4, 5

2, 3, 4

2, 3, 5

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Identifique la ecuación balanceada utilizando los números de oxidación


Cu = 1+, Cu = 0, Cu = 2+, H = 1+, S = 6+, S = 2-, O = 2-

4Cu + 4H2SO4 → Cu2S + 2CuSO4 + 4H2O

10Cu + 4H2SO43Cu2S + 4CuSO4 + 4H2O

5Cu + 4H2SO4 → Cu2S + 3CuSO4 + 4H2O

6Cu + 3H2SO43Cu2S + CuSO4 + 8H2O

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

¿Cuántos átomos de hidrógeno existen en 45 moléculas de C5H12O2?

12

45

225

540

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Identifique la relación que se debe utilizar para calcular el porcentaje de composición del hidrógeno en el compuesto metano (CH4), considerando que la masa molar de los elementos es C = 12 g/mol y H = 1 g/mol.

(116)100 %\left(\frac{1}{16}\right)\cdot100\ \%

(416)100 %\left(\frac{4}{16}\right)\cdot100\ \%

(516)100 %\left(\frac{5}{16}\right)\cdot100\ \%

(164)100 %\left(\frac{16}{4}\right)\cdot100\ \%

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