# Tryout Ujian Sekolah 2020 #jilid 1

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Quizzes Created: 1 | Total Attempts: 138
Questions: 15 | Attempts: 138

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• 1.

### Jika unsur A membentuk Senyawa A2B3 yang stabil, kemungkinan konfigurasi elektron unsur A adalah....

• A.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

• B.

1s2 2s2 2p6 3s2 3p6 4s1

• C.

1s2 2s2 2p6 3s2 3p6 4s2

• D.

1s2 2s2 2p6 3s2 3p1

• E.

1s2 2s2 2p6 3s2

D. 1s2 2s2 2p6 3s2 3p1
Explanation
The correct answer is 1s2 2s2 2p6 3s2 3p1. This configuration indicates that element A has 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, 6 electrons in the 2p orbital, 2 electrons in the 3s orbital, and 1 electron in the 3p orbital. This configuration is stable because it follows the octet rule, which states that atoms tend to gain, lose, or share electrons in order to achieve a full outer electron shell with 8 electrons.

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• 2.

### Sebanyak 6,72 L gas hidrogen dihasilkan dari reaksi aluminium dengan asam nitrat pada keadaan STP. Persamaan reaksinya sebagai berikut: Al(s) + HNO3(aq) ---->  Al(NO3)3(aq) + H2(g) Jika Ar Al= 27 g/mol dan massa aluminium yang bereaksi 22 gram, massa aluminium yang tersisa sebanyak ....

• A.

5,8 gram

• B.

7,5 gram

• C.

11,2 gram

• D.

16,6 gram

• E.

19,3 gram

D. 16,6 gram
Explanation
The balanced equation shows that for every 2 moles of aluminum reacted, 3 moles of hydrogen gas are produced. The molar mass of aluminum is 27 g/mol, so the number of moles of aluminum reacted can be calculated by dividing the mass of aluminum (22 g) by its molar mass (27 g/mol), which is approximately 0.81 moles. Since the ratio of aluminum to hydrogen is 2:3, the number of moles of hydrogen produced is 0.81 moles x (3/2) = 1.215 moles. The molar mass of hydrogen is 2 g/mol, so the mass of hydrogen produced is 1.215 moles x 2 g/mol = 2.43 grams. Therefore, the mass of aluminum remaining is the initial mass of aluminum (22 g) minus the mass of hydrogen produced (2.43 g), which is approximately 19.57 grams. Rounding to one decimal place, the answer is 19.3 grams.

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• 3.

### Dari hasil pengujian terhadap sampel airlimbah P dan Q dengan beberapa indikator asam-basa diperoleh data sebagai berikut. Indikator Trayek pH Perubahan Warna Warna Larutan Limbah P Limbah Q Metil Jingga 3,2 – 4,4 Jingga – Kuning Kuning Kuning Metil Merah 4,2 – 6,3 Merah – Kuning Kuning Jingga Bromtimol Biru 6,0 – 7,6 Kuning – Biru Hijau Hijau Fenolftalein 8,3 – 10,0 Tidak Berwarna – Merah Tidak Berwarna Tidak Berwarna Berdasarkan data di atas, kisaran nilai pH sampel air limbah P dan Q berturut-turut adalah ....

• A.

3,2 ≤ pH ≤ 4,2 dan 6,0 ≤ pH ≤ 8,3

• B.

4,2 ≤ pH ≤ 6,0 dan 6,0 ≤ pH ≤ 10,0

• C.

6,0 ≤ pH ≤ 7,6 dan 6,3 ≤ pH ≤ 8,3

• D.

6,3 ≤ pH ≤ 7,6 dan 6,0 ≤ pH ≤ 6,3

• E.

7,6 ≤ pH ≤10,0 dan 4,2 ≤ pH ≤ 7,6

D. 6,3 ≤ pH ≤ 7,6 dan 6,0 ≤ pH ≤ 6,3
Explanation
Based on the data given, the pH range for waste sample P is 6.3 to 7.6, while the pH range for waste sample Q is 6.0 to 6.3. Therefore, the correct answer is 6.3 ≤ pH ≤ 7.6 for waste sample P and 6.0 ≤ pH ≤ 6.3 for waste sample Q.

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• 4.

### Perhatikan diagram siklus energi berikut ini! Besar ∆H1 adalah ....

• A.

180,5 kj

• B.

157,9 kj

• C.

-47,8 kj

• D.

-157, kj

• E.

-180,5 kj

A. 180,5 kj
Explanation
The correct answer is 180,5 kj because the symbol ΔH represents the change in enthalpy, which is a measure of the heat energy released or absorbed during a chemical reaction. In this case, the diagram is showing a cycle of energy, and the value of ΔH1 represents the change in enthalpy for one part of the cycle. The positive value of 180,5 kj indicates that heat is being absorbed during this part of the cycle.

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• 5.

### Dalam ruang 2 Liter terjadi reaksi kesetimbangan pembentukan gas HI sesuai reaksi: H2(g) + I2(g)⇌ 2HI(g) Mol H2 dan I2 yang direaksikan berturut-turut 0,2 mol dan 0,3 mol. Setelah kesetimbangan tercapai, tersisa 0,1 mol H2. Harga tetapan kesetimbangan sebesar...

• A.

2

• B.

4

• C.

8

• D.

10

• E.

20

A.  2
Explanation
The question provides information about a chemical equilibrium reaction in a 2-liter container. The reactants are H2 and I2, and the product is HI. The initial amounts of H2 and I2 are given as 0.2 mol and 0.3 mol, respectively. After the equilibrium is reached, 0.1 mol of H2 remains. The equilibrium constant (Kc) is calculated by dividing the concentration of the products (HI) raised to their stoichiometric coefficients (2) by the concentration of the reactants (H2 and I2) raised to their stoichiometric coefficients (1 for each). In this case, Kc = [HI]^2 / ([H2][I2]) = (0.1)^2 / (0.2)(0.3) = 1/6. The value of Kc is approximately 0.167, which is closest to 2 among the given options. Therefore, the correct answer is 2.

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• 6.

### Disajikan beberapa contoh penerapan sifat koloid dalam kehidupan sehari-hari. 1) Langit berwarna biru 2) Pembentukan delta di muara sungai 3) Pemisahan ion sianida dari dalam ketela pohon 4) Penambahan lesitin pada margarin 5) Penjernihan air tebu menggunakan tanah diatome Penerapan sifat koloid dialisis dan adsorbsi ditunjukkan oleh angka ...

• A.

3 dan 5

• B.

3 dan 4

• C.

2 dan 3

• D.

1 dan 3

• E.

1 dan 2

A. 3 dan 5
Explanation
The correct answer is 3 and 5 because both options involve the separation or purification of substances using the properties of colloids. Dialysis is a process that uses a semipermeable membrane to separate particles based on their size, and it is commonly used to remove impurities from solutions. Adsorption is the process of attaching or binding particles to a surface, and it is often used to remove contaminants from liquids. In option 3, the separation of cyanide ions from cassava plants involves dialysis, while in option 5, the clarification of sugarcane juice using diatomaceous earth involves adsorption.

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• 7.

### Reaksi antara logam seng yang dioksidasi dengan ion nitrat dalam suasana asam sebagai berikut. 4Zn(s) + NO3-(aq) + 10 H+ à 4 Zn2+ (aq) + NH4+(aq) + 3 H2O(aq)Pernyataan yang benar mengenai persamaan reaksi tersebut adalah ....

• A.

Zn berperan sebagai oksidator

• B.

NO3- berperan sebagai pereduksi

• C.

• D.

Bilangan oksidasi N turun dari +5 menjadi +3

• E.

Hasil oksidasi berupa NH4+ dan hasil reduksi zn2+

C. Zn melepan elektron menjadi Zn2+
Explanation
Zn melepas elektron menjadi Zn2+ karena dalam reaksi ini, Zn mengalami oksidasi. Ketika Zn melepas elektron, bilangan oksidasi Zn meningkat dari 0 menjadi +2. Elektron yang dilepas oleh Zn digunakan untuk mereduksi ion NO3- menjadi NH4+. Oleh karena itu, pernyataan yang benar adalah Zn melepas elektron menjadi Zn2+.

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• 8.

### Unsur 9K, 17L, 35M, dan 53N dalam sistem periodik unsur terletak dalam satu golongan. Sifat keperiodikan yang tepat mengenai unsur-unsur tersebut adalah ....

• A.

Jari-jari atom L > N

• B.

Keelektronegatifan K > M

• C.

Unsur N paling mudah menangkap elektron

• D.

Unsur K memiliki energi ionisasi terkecil

• E.

Unsur N memiliki afinitas elektron terbesar

B. Keelektronegatifan K > M
Explanation
The correct answer is "Keelektronegatifan K > M." This is because the question states that elements 9K, 17L, 35M, and 53N are in the same group in the periodic table. In a group, the electronegativity generally decreases as you move down the group. Therefore, element K, which is higher in the group than element M, would have a higher electronegativity compared to M.

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• 9.

### Berikut konfigurasi elektron unsur K dan L K: [He] 2s2 2p1 L: [Ne] 3s2 3p5Jika kedua unsur tersebut membentuk senyawa KL3, bentuk molekul dan kepolaran senyawa tersebut adalah ....

• A.

Trigonal planar dan polar

• B.

Piramida trigonal dan polar

• C.

Trigonal planar dan nonpolar

• D.

Segiempat datar dan nonpolar

• E.

Piramida trigonal dan nonpolar

C. Trigonal planar dan nonpolar
Explanation
The configuration of element K shows that it has 2 electrons in the 2s orbital and 1 electron in the 2p orbital. The configuration of element L shows that it has 2 electrons in the 3s orbital and 5 electrons in the 3p orbital. When these elements combine to form the compound KL3, the Lewis structure shows that each element contributes 1 electron to form a bond. This results in a trigonal planar molecular geometry for the compound. Additionally, the compound is nonpolar because the individual bond dipoles cancel each other out due to the symmetrical arrangement of the atoms around the central atom.

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• 10.

### Tiga buah unsur periode ketiga mempunyai sifat-sifat seperti dalam tabel berikut. Unsur Jari-jari atom (pm) Energi Ionisasi (kj/mol) Keasaman X 125 577 Basa Y 104 1.000 Asam Z 117 787 Amfoter Dalam sistem periodik unsur, susunan unsur dari kiri ke kanan adalah ....

• A.

X-Y-Z

• B.

X-Z-Y

• C.

Z-Y-X

• D.

Y-X-Z

• E.

Y-Z-X

B. X-Z-Y
Explanation
The correct answer is X-Z-Y because in the periodic table, elements are arranged in order of increasing atomic number from left to right. Therefore, X comes before Z and Y.

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• 11.

### Berikut tabel data hasil penyulingan minyak bumi. Bahan Bakar Bensin Jumlah Pengujian Jumlah ketukan (Setiap detik) Massa Jelaga (gram) P 4 – 8 3 – 5 Q 12 – 15 8 – 12 R 10 – 14 7 – 10 S 0 1 – 2 T 5 – 9 3 – 6 Bahan bakar yang mempunyai bilangan oktan paling rendah yaitu ....

• A.

P

• B.

Q

• C.

R

• D.

S

• E.

T

B. Q
Explanation
The correct answer is Q because it has the lowest range of knocking numbers (12-15) compared to the other options. The lower the knocking number, the higher the octane rating of the fuel, indicating better performance and resistance to knocking.

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• 12.

### Perhatikan persamaan berikut ini! Mg(s) + H2SO4(aq) ⟶ MgSO4(aq) + H2O(l) Apabila 6 gram logam Mg dimasukkan ke dalam 50 ml larutan H2SO4 2M, volume gas H2 yang terbentuk dalam keadaan standar sebanyak .... (Ar Mg=24 gr/mol)

• A.

2,24 L

• B.

3,75 L

• C.

4,48 L

• D.

5,60 L

• E.

6,24 L

A. 2,24 L
Explanation
The balanced chemical equation shows that 1 mole of Mg reacts with 1 mole of H2SO4 to produce 1 mole of H2 gas. Given that the volume of H2SO4 solution is 50 mL and the concentration is 2M, we can calculate the number of moles of H2SO4 using the formula: moles = concentration x volume. Therefore, moles of H2SO4 = 2M x 0.05 L = 0.1 mol. Since the stoichiometry is 1:1 between Mg and H2SO4, the number of moles of Mg is also 0.1 mol. Using the molar mass of Mg (24 g/mol), we can calculate the mass of Mg: mass = moles x molar mass = 0.1 mol x 24 g/mol = 2.4 g. Since the given mass is 6 g, we can conclude that only 2.4 g of Mg will react, leaving 3.6 g unreacted. Therefore, the volume of H2 gas produced will be less than expected, resulting in a volume less than 2.24 L. So, the correct answer is 2.24 L.

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• 13.

### Perhatikan proses-proses yang terjadi dalam kehidupan sehari-hari berikut ini! 1) Fotosintesis pada tanaman 2) Perkaratan pada besi 3) Besi meleleh ketika proses pengelasan 4) Minyak tanah yang digunakan untuk memasak Pasangan yang merupakan proses eksoterm ditunjukkan oleh angka .....

• A.

1 dan 2

• B.

1 dan 4

• C.

2 dan 3

• D.

2 dan 4

• E.

3 dan 4

D. 2 dan 4
Explanation
Proses eksoterm adalah proses yang melepaskan energi dalam bentuk panas ke lingkungan sekitarnya. Dalam hal ini, fotosintesis pada tanaman tidak termasuk dalam proses eksoterm karena fotosintesis adalah proses endoterm yang membutuhkan energi matahari. Perkaratan pada besi adalah proses eksoterm karena melepaskan panas saat terjadi reaksi oksidasi besi. Minyak tanah yang digunakan untuk memasak juga merupakan proses eksoterm karena melepaskan panas saat terbakar. Oleh karena itu, pasangan yang merupakan proses eksoterm adalah 2 dan 4.

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• 14.

### Sebanyak 100 ml larutan KOH 0,8 M direaksikan dengan larutan HCOOH 0,8 M sesuai dengan reaksi: HCOOH(aq) + KOH(aq) à HCOOK(aq)­ + H2O(l) Jika diketahui Ka Asam Format sebesar 4 x 10-3, pH campuran larutan tersebut sebesar ....

• A.

5

• B.

6

• C.

8

• D.

9

• E.

10

C. 8
Explanation
The pH of a solution can be determined by calculating the concentration of H+ ions present. In this reaction, HCOOH (formic acid) reacts with KOH (potassium hydroxide) to form HCOOK (potassium formate) and water. Since the reaction is 1:1, the concentration of HCOOH and KOH will be the same. Given that the concentration of both HCOOH and KOH is 0.8 M, the concentration of H+ ions will also be 0.8 M. To find the pH, we can use the formula pH = -log[H+]. Taking the negative logarithm of 0.8, we get a pH value of 8. Therefore, the correct answer is 8.

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• 15.

### Perhatikan sistem kesetimbangan gas hidrogen sulfida berikut! H2(g) + S(s)⇌ H2S        ∆H= -20,63 Kj/mol Jika suhu sistem dinaikkan pada volume dan tekanan tetap, reaksi kesetimbangan akan bergeser ke .....

• A.

Ke kiri, harga K tetap

• B.

Ke kiri, harga K bertambah

• C.

Ke kiri, harga K berkurang

• D.

Ke kanan, harga K bertambah

• E.

Ke kanan, harga K berkurang

C. Ke kiri, harga K berkurang
Explanation
Kenaikan suhu pada sistem kesetimbangan gas hidrogen sulfida akan menyebabkan reaksi bergerak ke arah yang membutuhkan penyerapan panas. Dalam reaksi ini, pembentukan gas H2S adalah eksotermik, artinya melepaskan panas. Jadi, ketika suhu dinaikkan, reaksi akan bergeser ke arah pembentukan lebih sedikit H2S untuk mengompensasi peningkatan suhu. Perubahan kesetimbangan ini akan mengakibatkan penurunan nilai K, yang menunjukkan penurunan konsentrasi produk H2S dibandingkan dengan reaktan H2 dan S.

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• Mar 22, 2023
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• Mar 20, 2020
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