Question

PLS TELL THE ANS FAST I WILL MARK AS BRANLIEST

Answer 1

Answer:

Final mode (2a−2a)

Initial mole 2a

2H

2

(g)

+

(a−x)

a

O

2(g)

→

2x

0

2H

2

O

(g)

Given 2x=2a×

100

80

=1.6a

∴x=0.8a

Thus, after the reaction H

2

left=2a−1.6a=0.4a mole

O

2

left=0.2a mole

H

2

O formed=1.6a mole

∴ Total mole at 120

∘

C in gaseous phase

=0.4a+0.2a+1.6a=2.2a

Now, given at initial conditions P=0.8 atm,T=293 K

P×V=nRT

0.8 V=3a×R×293

∴V=

0.8

3a×R×293

The volume of container remains constant.

After the reaction using, P×V=nRT

∴P×

0.8

3a×R×293

=2.2a×R×393

P=

3×293

393×0.8×2.2

=0.787 atm

Answer 2

Sin90=1

cos45=1

tan 45=1

2(1) is a answer