1. The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons.2.CaH2(s) + H2O(1) Ca(OH)2(s) + H2(g) (not balanced)(a) How many grams of H2(g) result from the reac-tion of 127 g CaH2 with an excess of water?(b) How many grams of water are consumed in the reaction of 56.2 g CaH2?(c) What mass of CaH2(s) must react with an excess of water to produce 8.12 × 1024 molecules of H₂?A side reaction in the manufacture of rayon from wood pulp is 3 CS2 + 6 NaOH→ 2 Na2CS3 + Na2CO3 + 3 H2OHow many grams of Na2CS3 are produced in the reac-tion of 92.5 mL of liquid CS2 (d = 1.26 g/mL) and 2.78 mol NaOH?3. Titanium tetrachloride, TiCl, is prepared by the reac-tion below.3 TiO2(s) + 4 C(s) + 6 Cl₂(g) 3 TiCl4(g) + 2 CO2(g) + 2 CO(g)What is the maximum mass of TiCl, that can be obtained from 35 g TiO2, 45 g Cl₂, and 11 g C?4. In the reaction shown, 100.0 g C6H11OH yielded 64.0 g C6H10. (a) What is the theoretical yield of the reaction?(b) What is the percent yield? (c) What mass ofC6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)? C6H11OH→ C6H10 + H2O5. Azobenzene, an intermediate in the manufacture of dyes, can be prepared from nitrobenzene by reaction with triethylene glycol in the presence of Zn and KOH. In one reaction, 0.10 L of nitrobenzene (d 1.20 g/mL) and 0.30 L of triethylene glycol (d 1.12 g/mL) yields 55 g azobenzene. What are the (a) theoretical yield, (b) actual yield, and (c) percent yield of this reaction? 2 C6H5NO2 + 4 C6H14O4 nitrobenzene triethylene glycol Zn KOH (C6H5N)2+4C6H12O4+4 H₂O azobenzene
Answer (4)
In fresh water sound waves travel at 1497m/s at 25 degrees, I'll assume that's the characteristics of the water.
If it's 0.01s then you need to divide the speed by 100 to get the, 14.97, however it gets there and back in that time so you need to halve it. 7.485m
The fish are approximately 15 meters away from the place where the sound pulse was emitted. ;
The school of fish is approximately 7.5 meters away from the point where the sound pulse was emitted. This is calculated using the speed of sound in water and the time it took for the pulse to travel to the fish and back. By dividing the total time by two and applying the distance formula, we find the distance to the fish.
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Jawaban:Soal 1:Reaksi kalsium hidrida dengan air menghasilkan gas hidrogen.CaH2(s) + H2O(l) -> Ca(OH)2(s) + H2(g) (belum setara) (a) Berapa gram H2(g) yang dihasilkan dari reaksi 127 g CaH2 dengan kelebihan air? Langkah 1: Setarakan Persamaan Reaksi Persamaan setara: CaH2(s) + 2 H2O(l) -> Ca(OH)2(s) + 2 H2(g) Langkah 2: Hitung Massa Molar CaH2 Massa molar CaH2 = 40 (Ca) + 2(1) (H) = 42 g/mol Langkah 3: Hitung Mol CaH2 Mol CaH2 = massa / massa molar = 127 g / 42 g/mol = 3.02 mol Langkah 4: Gunakan Stoikiometri untuk Menghitung Mol H2 Dari persamaan setara, 1 mol CaH2 menghasilkan 2 mol H2. Mol H2 = 2 * mol CaH2 = 2 * 3.02 mol = 6.04 mol Langkah 5: Hitung Massa H2 Massa molar H2 = 2 g/mol Massa H2 = mol H2 * massa molar H2 = 6.04 mol * 2 g/mol = 12.08 g Jawaban (a): 1. 08 gram H2 dihasilkan. (b) Berapa gram air yang dikonsumsi dalam reaksi 56.2 g CaH2? Langkah 1: Hitung Mol CaH2 Mol CaH2 = massa / massa molar = 56.2 g / 42 g/mol = 1.34 mol Langkah 2: Gunakan Stoikiometri untuk Menghitung Mol H2O Dari persamaan setara, 1 mol CaH2 bereaksi dengan 2 mol H2O. Mol H2O = 2 * mol CaH2 = 2 * 1.34 mol = 2.68 mol Langkah 3: Hitung Massa H2O Massa molar H2O = 18 g/mol Massa H2O = mol H2O * massa molar H2O = 2.68 mol * 18 g/mol = 48.24 g Jawaban (b):1. 24 gram air dikonsumsi. (c) Berapa massa CaH2(s) yang harus bereaksi dengan kelebihan air untuk menghasilkan 8.12 x 10^24 molekul H2?Langkah 1: Hitung Mol H21 mol = 6.022 x 10^23 molekulMol H2 = (8.12 x 10^24 molekul) / (6.022 x 10^23 molekul/mol) = 13.48 molLangkah 2: Gunakan Stoikiometri untuk Menghitung Mol CaH2Dari persamaan setara, 2 mol H2 dihasilkan dari 1 mol CaH2.Mol CaH2 = mol H2 / 2 = 13.48 mol / 2 = 6.74 molLangkah 3: Hitung Massa CaH2Massa CaH2 = mol CaH2 * massa molar CaH2 = 6.74 mol * 42 g/mol = 283.08 gJawaban (c):1. 08 gram CaH2 harus bereaksi. Soal 2:Reaksi samping dalam pembuatan rayon dari pulp kayu:3 CS2 + 6 NaOH -> 2 Na2CS3 + Na2CO3 + 3 H2OBerapa gram Na2CS3 yang dihasilkan dalam reaksi 92.5 mL CS2 (d = 1.26 g/mL) dan 2.78 mol NaOH?Langkah 1: Hitung Massa CS2Massa CS2 = volume * densitas = 92.5 mL * 1.26 g/mL = 116.55 gLangkah 2: Hitung Mol CS2Massa molar CS2 = 12 (C) + 2(32) (S) = 76 g/molMol CS2 = massa / massa molar = 116.55 g / 76 g/mol = 1.53 molLangkah 3: Tentukan Pereaksi PembatasDari persamaan setara, 3 mol CS2 bereaksi dengan 6 mol NaOH (rasio 1:2).Kita punya 1.53 mol CS2 dan 2.78 mol NaOH.Jika semua CS2 bereaksi, kita butuh 1.53 mol * 2 = 3.06 mol NaOH. Kita hanya punya 2.78 mol NaOH, jadi NaOH adalah pereaksi pembatas.Langkah 4: Gunakan Stoikiometri untuk Menghitung Mol Na2CS3Dari persamaan setara, 6 mol NaOH menghasilkan 2 mol Na2CS3 (rasio 3:1).Mol Na2CS3 = (mol NaOH / 6) * 2 = (2.78 mol / 6) * 2 = 0.93 molLangkah 5: Hitung Massa Na2CS3Massa molar Na2CS3 = 2(23) (Na) + 12 (C) + 3(32) (S) = 154 g/molMassa Na2CS3 = mol Na2CS3 * massa molar Na2CS3 = 0.93 mol * 154 g/mol = 143.22 gJawaban:1. 22 gram Na2CS3 dihasilkan.Penjelasan:Soal 3: Titanium tetraklorida, TiCl4, dibuat melalui reaksi: 3 TiO2(s) + 4 C(s) + 6 Cl2(g) -> 3 TiCl4(g) + 2 CO2(g) + 2 CO(g) Berapa massa maksimum TiCl4 yang dapat diperoleh dari 35 g TiO2, 45 g Cl2, dan 11 g C? Langkah 1: Hitung Massa Molar Setiap Pereaksi dan Produk - TiO2: 47.87 (Ti) + 2 * 16 (O) = 79.87 g/mol- C: 12.01 g/mol- Cl2: 2 * 35.45 (Cl) = 70.90 g/mol- TiCl4: 47.87 (Ti) + 4 * 35.45 (Cl) = 189.67 g/mol Langkah 2: Hitung Mol Setiap Pereaksi - TiO2: 35 g / 79.87 g/mol = 0.438 mol- Cl2: 45 g / 70.90 g/mol = 0.635 mol- C: 11 g / 12.01 g/mol = 0.916 mol Langkah 3: Tentukan Pereaksi Pembatas Kita perlu membandingkan rasio mol pereaksi dengan rasio stoikiometri dalam persamaan reaksi. - TiO2: Dibutuhkan 3 mol TiO2- C: Dibutuhkan 4 mol C- Cl2: Dibutuhkan 6 mol Cl2 Berdasarkan mol yang kita miliki: - TiO2: 0.438 mol / 3 = 0.146- C: 0.916 mol / 4 = 0.229- Cl2: 0.635 mol / 6 = 0.106 Cl2 memiliki nilai terkecil, jadi Cl2 adalah pereaksi pembatas. Langkah 4: Hitung Mol TiCl4 yang Dapat Dihasilkan Dari persamaan setara, 6 mol Cl2 menghasilkan 3 mol TiCl4. Mol TiCl4 = (mol Cl2 / 6) * 3 = (0.635 mol / 6) * 3 = 0.3175 mol Langkah 5: Hitung Massa TiCl4 yang Dapat Dihasilkan Massa TiCl4 = mol TiCl4 * massa molar TiCl4 = 0.3175 mol * 189.67 g/mol = 60.22 g Jawaban: Massa maksimum TiCl4 yang dapat diperoleh adalah 60.22 gram. maaf cuman bisa bantu jawab segitu,kolom jawabnya tidak muat,penuh
