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Irina-Kira [14]
3 years ago
8

HELPP

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
4 0
I’m pretty sure it’s D.increases the activation energy for a reaction.
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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th
Black_prince [1.1K]

Answer:

K_{goal}=1.793*10^{-33}

Explanation:

N2(g)+O2(g)⇌2NO(g), K_1 = 4.10*10^{-31}

N2(g)+2H2(g)⇌N2H4(g), K_2 = 7.40*10^{-26}

2H2O(g)⇌2H2(g)+O2(g), K_3 = 1.06*10^{-10}

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g)                     Eq (1)

Equilibrium constant for Eq (1) is K_1*K_3*K_3

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g)                       Eq (2)  

Equilibrium constant for Eq (2) is (K_1*K_3*K_3)^{1/2}

Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}

7 0
3 years ago
How many Helium atoms are in 12 grams of Helium?
horsena [70]
12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms
6 0
3 years ago
For the reaction, Cl2 + 2KBr --> 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b
Sedaia [141]

Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

5 0
3 years ago
Is 2Zn + 2HCI = ZnCL2 + H2 balanced or unbalanced
riadik2000 [5.3K]
Unbalanced it should be 2Zn+2Hcl=2ZnCl2+H2
7 0
3 years ago
A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau
spayn [35]

Answer:

C₅ H₁₂ O

Explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of  H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound =  8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O

6 0
3 years ago
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