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6. Tentukan dimensi konstanta gravitasi umum !
Jawab :
Dimensi konstanta gravitasi umum
Gaya  F = m . a
= kg . m/s2
= M . L/T2 = M x L x T-2
= [M] [L] [T]-2
G = F x r2
m1x m2
= [M] [L] [T]-2 x [L]2
[M]2
= [M] [M]-2 [L]3 [T]-2
= [M]-1 [L]3 [T]-2
7. 2 pegas identik dengan konstanta gaya 800 N/m kedua pegas dipararelkan hitung h, gaya yang dibutuhkan untuk menarik pegas sehingga bertambah panjang 5 cm.
Jawab :
D1 K = 800 N/m x 2
= 1600 N/m
Λx = 5 cm = 5 – 10-2 m
D2 F = ….?
D3 F = K . Λx
= 1600 . 5 . 10-2
= 80 Newton
8. Tuliskan Dimensi
a) Konstanta Pegas!
b) Tegangan!
Jawab :
a) Konstanta Pegas (N/m)
N/m = Nm-1 = kg m s-2 m-1
= kg . s-2
= [M] [T]-2
b) Tegangan (N/m2)
N/m-2 = kg . m . s-2 m-2
= kg . m-1 . s-2
= [M] [L]-1 [T]-2
9. Dua buah planet A dan B mengorbit mengitari matahari. Perbandingan antar planet A ke B ke matahari adalah RA : RB = 1 : 4. Apabila periode planet A mengelilingi matahari adalah 88 hari. Tentukan periode planet B mengelilingi matahari!
Jawab :
D1 RA : RB = 1 :4
T1 = 88 hari
D2 T2 = ….?
D3 T12 = R13
T22 R23
882 = 13
T22 43
T22 = 882 x 43
T22 = 495616

T2 = V495616
T2 = 704 hari
10. Sebuah partikel melakukan gerak harmonic sederhana dengan periode ¼ sekon dan amplitude 2 cm. hitung kecepatan gerak maksimum!
Jawab :
D1 T = ¼ s
A = 2 cm
D2 Vy max …?
D3 V max = W . A  W = 2π/T
= 8π . 2 = 2π/¼
= 16 π cm/s = 8π

Entalpi pembentukan H2O = -242 Kj / mol

Energi ikatan H – H = 436 Kj/mol

Energi ikatan O = O = 495 Kj/mol

Energi ikatan rata – rata O – H dalam air adalah……..

Jawab :

2H2 + O2 → 2H2O

2 (H – H) + O = O → 2 (H – O – H )

Reaktan:

2 (H – H ) = 2 . 436 = 872 Kj/ mol

O = O = 495 Kj/ mol

Jml = 1367 Kj/mol

Produk:

4 . O – H = ….

∆H = ∑E Reaktan – ∑EProduk

-242 = 1367 – 4 (O – H )

4 ( O – H ) = 1367 + 242

4 ( O – H ) = 1609

( O – H ) = 1609 / 4

( O – H ) = 402,25 Kj / mol

Diketahui : 2C + 3H2 → C2H6 ∆H = -85 Kj/mol

Energi ikatan C – C = 349 Kj/mol

Energi ikatan H – H = 436 Kj/mol

Energi penguapan C = 718 Kj/mol

Tentukan energy ikatan rata – rata C – H pada C2H6 !

Jawab :
2C + 3H2 → C2H6 ∆H = -85 Kj/mol

H H

2C + 3(H – H) → H – C – C – H

H H

Reaktan:

2 . C = 2 . 718 = 1436

3 (H – H ) = 3 . 436 = 1.308

Jumlah : 2744 Kj/mol

Produk:

6 (C – H ) = …

C – C = 349 Kj/mol

∆H = ∑E Reaktan – ∑EProduk

-85 = 2744 – (349 + 6 . C – H )

349 + 6 .( C – H ) = 2744 + 85

349 + 6 . ( C – H ) = 2829

6 . ( C – H ) = 2829 – 349

6 . ( C – H ) = 2480

C – H = 2480 / 6

C – H = 413,3 Kj/mol

Diketahiu :

C = C = 614 KJ/Mol C – Cl = 328 KJ/Mol

C – C = 348 KJ/Mol H – Cl = 431 KJ/Mol

C – C = 413 KJ/Mol

∆Hr dari reaksi H2C=CH2 + HCl→ H2C – CH2Cl adalah….

Jawab…

H2C = CH2 + HCl → H2C – CH2 Cl

H – C = C –H + H – Cl → H – C – C – Cl

H H H H

R → C – H x 4 = 413 x 4 = 1652

C = C = = 614

H – Cl = = 431 +

= 2697

P → C – H x 4 = 413 x 4 = 1652

C – C = = 348

C – Cl = = 328 +

= 2328

∆Hr = ∑ ER – ∑ EP

= 2697 – 2328

= 369

Diketahiu ∆HOf NH2(g) = – 46 KJ/Mol.Energi ikatan

N = N = 945 KJ/Mol,H – H = 436 KJ/Mol.Hitunglah energy ikatan rata-rata NH!

Jawab :

D1 : ∆HO NH3 = -46 KJ/Mol

N Ξ N = 945 KJ/Mol

H – H = 435 KJ/Mol

D2 : H – N …..?

D3 : 3H2 + N2 → 2 NH3

3( H – H ) + N Ξ N → 2 H – N – H

∆H = ∑ER – ∑EP

-46 = 2250 – 6 ( N – H )

6 ( N – H ) = 2250 + 46

6 ( N – H ) = 2296

( N – H ) = 382,67 KJ/Mol

REAKTAN :

3 . ( H – H ) = 3 . 435 = 1305

N Ξ N = 945 +

=2250

PRODUK :

2 . 3 ( H – H ) = 6 ( H – H )

Jika energy ikatan rata-rata

C – C = 346 KJ/Mol C = C =598 KJ/Mol H – H = 436 KJ/Mol

C = C = 346 KJ/Mol C – H = 415 KJ/Mol

Perubahan entalpi untuk reaksi C2H4 + H2 → C2H8 Adalah….

JAWAB :

H H

H – C = C – H + H – H → H – C – C – H

H H H H

REAKTAN :

C – H x 4 = 415 x4 = 1660

C = C = 598

H – H = 436 +

= 2694

PRODUK :

C – H x 6 = 415 x 6 = 2490

C – C = 346 +

= 2836

∆H = ∑ER – ∑EP

= 2694 – 2836

= – 142 KJ/Mol

Diketahui data energy ikatan sebagai berikut (Ar C=12,H=1)

C – H = 410KJ/Mol

C = O =732 KJ/Mol

O – H = 460KJ/Mol

O = O = 489KJ/Mol

C Ξ C = 828KJ/Mol

Perubahan dan entalpi pada pembakaran 26 gram gas C2H2 menurut reaksi :

C2H2(g) + 2,5O2(g)→2CO2(g) + H2O(g) adalah….

JAWAB :

H – C Ξ C – H + 2,5 (O = O)→2 (O = C = O ) + H – O – H

REAKTAN :

C – H x2 = 410 x 2 = 820

C = C = 828

O = O x 2,5 = 489 x 2,5 =1222,5 +

=2870,5

PRODUK :

2 ( C = O x 2) = 732 x 4 = 2928

H – O x 2 = 460 x 2 = 920 +

= 3848

∆H = ∑ER – ∑EP

= 2870,5 – 3848

` = – 977,5 KJ/Mol

Mr C2H2 = ( 12 x 2 ) + ( 2 x 1 )

=24 + 2

= 26

Mol C2H2 = gr/Mr

= 26/26

= 1 Mol

∆H pembakaran 26 kg C2H2 adalah = Mol x ∆H

= 1 x (-977,5)

=-977,5

Diketahui energy ikat rata-rata sebagai berikut :

H – H = 104,2 KKal/mol

Cl – Cl = 57,8 KKal/mol

H – Cl = 103,1 KKal/mol

Kalor yang diperlukan untuk menguraikan 182,5 gr HCl (ArH=1,Cl=35,5)menjadi unsure-unsurnya adalah….

JAWAB

2HCl→H2 + Cl2

2 H – Cl →H – H + Cl – Cl

REAKTAN :

2 H – Cl = 2 x 103 x 1

=206

PRODUK :

H – H = 104,2

Cl – Cl = 58,8 +

= 162

∆H =∑ER – ∑EP

= 206,2 – 162

=44,2

Mol HCl = gr/Mr → Mr HCl = 36,5

= 182,5/36,5

= 5

Kalor yang diperlukan untuk menguraikan 182,5 gr HCl menjadi unsur-unsurnya adalah…

∆H x HCl = 44,2 x 5

= 221 KKal

CH3 – CH2OH + O2 → CH3 – C

H H H

H – C – C – O – H + O = O → H – C – C + H – O – H

O – H

H H H

REAKTAN :

C – H x 5 = 414 x 5 = 2070

C – C = 374

C – O = 360

O – H = 646

O = O = 736 +

=4186

PRODUK :

C – Hx 3 = 414 x 3 = 1242

C – C = 374

C = O = 736

C – O = 360

O – H x 3 = 646 x 3 = 1938 +

= 4650

∆H = ∑ER – ∑EP

= 4186 – 4650

= -464 KJ/mol

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