REMIDI KIMIA KLAS XI SMA SHAFTA SBY
Reaksi akan berlangsung 3 kali lebih cepat dari semula setiap kenaikan 10 derajat C. Jika pada
suhu 40 derajat C suatu reaksi berlangsung 3 menit maka pada suhu 70 derajat C reaksi akan berlangsung berapa menit ?
Jawaban paling lambat tanggal 31 januari jam 9 pagi
posted by dwi agus dh at
7:43 PM
14 Comments:
diket: suhu1:40derajat C
suhu2:70 derajt C
t1 :3 menit= 180 sekon
dita :t2...?
jawab:
delta t=(70-40)=30 derajat C
v1= 1/t1=1/180
v2= 2(30/10)x1/180
= 2x2x2x1/180
= 8x1/180
= 2/45
t2 =1/v2
=1:2/45
=45/2 = 22,5 sekon
jadi pada suhu 70 derajat C reaksi akan berlangsung 22,5 sekon
January 28, 2008 at 9:16 PM
diket: suhu1:40derajat C
suhu2:70 derajt C
t1 :3 menit= 180 sekon
dita :t2...?
jawab:
delta t=(70-40)=30 derajat C
v1= 1/t1=1/180
v2= 2(30/10)x1/180
= 2x2x2x1/180
= 8x1/180
= 2/45
t2 =1/v2
=1:2/45
=45/2 = 22,5 sekon
jadi pada suhu 70 derajat C reaksi akan berlangsung 22,5 sekon
January 28, 2008 at 9:16 PM
diket: suhu1:40derajat C
suhu2:70 derajt C
t1 :3 menit= 180 sekon
dita :t2...?
jawab:
delta t=(70-40)=30 derajat C
v1= 1/t1=1/180
v2= 2(30/10)x1/180
= 2x2x2x1/180
= 8x1/180
= 2/45
t2 =1/v2
=1:2/45
=45/2 = 22,5 sekon
jadi pada suhu 70 derajat C reaksi akan berlangsung 22,5 sekon
January 28, 2008 at 9:16 PM
dit :t2...?
70-40
jawab :t2 = (1/3)._____ .t1
10
3
=(1/3) .3
= 1/27 .3
= 1/9 menit
January 29, 2008 at 9:10 PM
dit :t2...?
70-40
jawab :t2 = (1/3)._____ .t1
10
3
=(1/3) .3
= 1/27 .3
= 1/9 menit
January 29, 2008 at 9:15 PM
MAYTA TRIYANTI XI.IPA SMA SHAFTA
Dit:t2...?
70-40
jawab : t2 = (1/3).-----.t1
10
3
= (1/3) . 3
= 1/27 . 3
= 1/9 menit
January 29, 2008 at 9:26 PM
dit; t2.../
70-40
jawab : t2 = (1/3).----- . t1
10
3
= (1/3) . 3
= 1/27 . 3
= 1/9 menit
January 29, 2008 at 9:31 PM
dit : t2....?
70-40
jawab ; t2 = (1/3) . ----- .t1
10
3
=(1/3) . 3
=1/27 . 3
= 1/9 menit
January 29, 2008 at 9:36 PM
diket:suhu1=40derajat c
suhu 2=70derajat c
ti=3menit=180derajat c
dit: t2
jawab : t =(70-40)
v1=1/t1=1/180
v2=2/(30/10)x1)180
=2x2x2x1/180
= 8x1/180
=2/45
=1:2/45
=45/2=2,25s
nama arum grace
xi-ipa/04
January 30, 2008 at 4:57 AM
diket:suhu1=40derajat c
suhu 2=70derajat c
ti=3menit=180derajat c
dit: t2
jawab : t =(70-40)
v1=1/t1=1/180
v2=2/(30/10)x1)180
=2x2x2x1/180
= 8x1/180
=2/45
=1:2/45
=45/2=2,25s
nama arum grace
xi-ipa/04
January 30, 2008 at 4:57 AM
FENTI NURFIA/XI-IPA
Jawab:
Δt = (70-40)˚C = 30˚C
V2 = 3(pangkat30/10) x 1/3
= (3x3x3)x 1/3
= 1/9
t2 = 1/v = 1:1/9
= 9 menit
Jadi, lamanya reaksi pada temperature 70˚C adalah 9 menit.
January 30, 2008 at 5:28 AM
diket: suhu1:40derajat C
suhu2:70 derajt C
t1 :3 menit= 180 sekon
dita :t2...?
jawab:
delta t=(70-40)=30 derajat C
v1= 1/t1=1/180
v2= 2(30/10)x1/180
= 2x2x2x1/180
= 8x1/180
= 2/45
t2 =1/v2
=1:2/45
=45/2 = 22,5 sekon
jadi pada suhu 70 derajat C reaksi akan berlangsung 22,5 sek
bagus p (x1_ipa)
February 2, 2008 at 8:46 PM
diket:suhu1:40derajat c
suhu2:70 derajat c
t1:3 menit= 180 sekon
dit:t2...?
jawab:
delta t=(70-40)=30 derajat c
v1=1/t1=1/180
v2=2(30/10)*1/180
=2x2x2x1/180
=8x1/180
=1:2/45
t2=1/v2
1:2/45
=45/2=22,5 sekon
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February 3, 2008 at 3:30 AM
PAK AGUS, Q SYAHRUL (NOMO) SHAFTA . INTERNETX DAH KONEK KE LAB. KOMPTER & RUANG MMEDIA
TRUZ KITA NDIRIKAN KOMUNITAS IT DI SMA SHAFTA YANG...............
SAMA PK ANTON.
UDAH PAK YACH Q MAU FORMAT ULANG SEMUANYA
September 15, 2008 at 8:39 PM
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