Iklan

Iklan

Pertanyaan

D a t a d a r i k ese t imban g an : 2 H I + 4 m u ( g ) ⇆ H 2 ​ ( g ) + I 2 ​ ( g ) H a r g a t e t a p an k ese t imban g an ( Kc ) d a r i re ak s i t erse b u t a d a l ah ...

  1. K c space equals space fraction numerator open square brackets 0 comma 02 close square brackets space open square brackets 0 comma 02 close square brackets over denominator open square brackets 0 comma 06 close square brackets end fraction

  2. K c space equals space fraction numerator open square brackets 0 comma 02 close square brackets space open square brackets 0 comma 02 close square brackets over denominator open square brackets 0 comma 06 close square brackets squared end fraction

  3. K c space equals space fraction numerator open square brackets 0 comma 06 close square brackets squared over denominator open square brackets 0 comma 02 close square brackets space open square brackets 0 comma 02 close square brackets end fraction

  4. K c space equals space fraction numerator open square brackets 0 comma 04 close square brackets space open square brackets 0 comma 02 close square brackets over denominator open square brackets 0 comma 02 close square brackets space end fraction

  5. K c space equals space fraction numerator open square brackets 0 comma 04 close square brackets over denominator open square brackets 0 comma 02 close square brackets space open square brackets 0 comma 02 close square brackets end fraction

Iklan

A. Ratna

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Iklan

Pembahasan

K c space equals space fraction numerator left square bracket I subscript 2 right square bracket space left square bracket H subscript 2 right square bracket over denominator left square bracket H I right square bracket squared end fraction space equals space fraction numerator left square bracket 0 comma 02 right square bracket space left square bracket 0 comma 02 right square bracket over denominator left square bracket 0 comma 06 right square bracket squared end fraction

Latihan Bab

Pendahuluan Kesetimbangan Kimia

Konsep Kesetimbangan Kimia

Konstanta Kesetimbangan (K)

Konstanta Kesetimbangan Konsentrasi (Kc)

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

14

Iklan

Iklan

Pertanyaan serupa

Dalam bejana yang volumenya 5 liter dipanaskan gas PCl 5 ​ sebanyak 0,4 mol pada suhu 400 K sehingga terjadi reaksi kesetimbangan: PCl 5 ​ ( g ) ⇌ PCl 3 ​ ( g ) + Cl 2 ​ ( g ) Pada saat setim...

2rb+

4.2

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Fitur Roboguru

Topik Roboguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2023 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia