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The New Left was a political movement that was largely led by youth in the late 1950s into the 1960s. The New Left was a movement that was built out of estrangement of young people during this period with their government and the cultural constraints of their society. The New Left grew throughout the 1960s as politically left movements grew around the world in response to revolutionary fervor. Specifically in the US there was growing anger and backlash against the government because of the continuation and expansion of the War in Vietnam. The draft and the war only led to an increase in the New Left movement and also the anti-war movement. The movement ended up fracturing and becoming overshadowed because of violent activity of some of its members that eventually led to the movements exit from politics.

Answered by Gibbs | 2024-06-10

The New Left rose in the late 1950s and 1960s due to youth dissatisfaction with the government, the civil rights movement, and opposition to the Vietnam War. The war and draft mobilized young people, fostering coalitions and leading to significant cultural and political changes in American society, although the movement eventually fragmented by the late 1970s. This period marked a transition toward more radical reforms and influenced later political developments in the U.S.
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Answered by Gibbs | 2024-10-31

Jawaban:2. Tentukan zat yang bertindak sebagai oksidator dan reduktor!| No | Reaksi | Oksidator | Reduktor ||----|--------------------------------------------------------------------|------------------|--------------|| 1 | Na(s) + Cl₂(g) → NaCl(aq) | Cl₂ | Na || 2 | Fe²⁺ + MnO₄⁻ + H⁺ → Mn²⁺ + Fe³⁺ | MnO₄⁻ | Fe²⁺ || 3 | 2HCl + SnCl₂ → 2FeCl₃ + SnCl₄ *(asumsi reaksi benar)* | FeCl₃ | SnCl₂ || 4 | Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) | H⁺ (dari H₂SO₄) | Zn || 5 | 2H₂O₂(l) → 2H₂O(l) + O₂(g) | H₂O₂ | H₂O₂ |Catatan:- Oksidator: Zat yang mengalami reduksi (menurunkan biloks, menerima e⁻)- Reduktor: Zat yang mengalami oksidasi (menaikkan biloks, melepas e⁻)-----------------------------------------------------------3. Setarakan reaksi redoks berikut ini dengan metode setengah reaksi (suasana asam):H₂S(g) + Cr₂O₇²⁻(aq) → S(s) + Cr³⁺(aq)Langkah-langkah:a. Reaksi setengah oksidasi: H₂S → S + 2e⁻b. Reaksi setengah reduksi: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂Oc. Samakan jumlah elektron: 3(H₂S → S + 2e⁻) 1(Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O)d. Jumlahkan: 3H₂S + Cr₂O₇²⁻ + 14H⁺ → 3S + 2Cr³⁺ + 7H₂O-----------------------------------------------------------4. Setarakan reaksi redoks berikut ini dengan metode bilangan oksidasi (suasana basa):Zn(s) + NO₃⁻(aq) → ZnO₂²⁻(aq) + NH₃(g)Langkah-langkah:- Zn: 0 → +2 (Naik 2 → Oksidasi)- N (NO₃⁻): +5 → -3 (Turun 8 → Reduksi)a. KPK e⁻ = 8 Zn dikali 4 (4 × 2 = 8 e⁻) NO₃⁻ dikali 1b. Tambahkan H₂O untuk O, OH⁻ untuk basa, H₂O untuk H jika perluHasil setara akhir (suasana basa):4Zn + NO₃⁻ + 7H₂O → 4ZnO₂²⁻ + NH₃ + 2OH⁻Catatan:- Pastikan jumlah atom dan muatan setara di kiri dan kanan.

Answered by pupusdwisuryoanggono | 2025-07-20